Tangents to parabola $y^{2}=4ax$ meet hyperbola $x^2/a^2-y^2/b^2=1$ at $A$ and $B$. Find the locus of...
$begingroup$
If tangents to the parabola $y^{2} = 4ax$ intersect the hyperbola $dfrac{x^2}{a^2} - dfrac{y^2}{b^2} = 1$ at $A$ and $B$, then find the locus of point of intersection of tangents at $A$ and $B$.
I know that tangent to parabola is $y = mx + a/m$ ($m$ being the slope), but I am not able to figure out how to take out point of intersections.
analytic-geometry conic-sections tangent-line
edited Dec 29 '18 at 23:30
Blue
47.7k870151
asked Dec 29 '18 at 15:56
BadguyBadguy
63
$endgroup$
add a comment |
$begingroup$
If tangents to the parabola $y^{2} = 4ax$ intersect the hyperbola $dfrac{x^2}{a^2} - dfrac{y^2}{b^2} = 1$ at $A$ and $B$, then find the locus of point of intersection of tangents at $A$ and $B$.
I know that tangent to parabola is $y = mx + a/m$ ($m$ being the slope), but I am not able to figure out how to take out point of intersections.
analytic-geometry conic-sections tangent-line
edited Dec 29 '18 at 23:30
Blue
47.7k870151
asked Dec 29 '18 at 15:56
BadguyBadguy
63
$endgroup$
1
$begingroup$
Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
$endgroup$
– Blue
Dec 29 '18 at 23:30
$begingroup$
I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
$endgroup$
– Badguy
Dec 30 '18 at 8:47
$begingroup$
What is "$T$" ?
$endgroup$
– Blue
Dec 30 '18 at 8:50
add a comment |
$begingroup$
If tangents to the parabola $y^{2} = 4ax$ intersect the hyperbola $dfrac{x^2}{a^2} - dfrac{y^2}{b^2} = 1$ at $A$ and $B$, then find the locus of point of intersection of tangents at $A$ and $B$.
I know that tangent to parabola is $y = mx + a/m$ ($m$ being the slope), but I am not able to figure out how to take out point of intersections.
analytic-geometry conic-sections tangent-line
edited Dec 29 '18 at 23:30
Blue
47.7k870151
asked Dec 29 '18 at 15:56
BadguyBadguy
63
$endgroup$
If tangents to the parabola $y^{2} = 4ax$ intersect the hyperbola $dfrac{x^2}{a^2} - dfrac{y^2}{b^2} = 1$ at $A$ and $B$, then find the locus of point of intersection of tangents at $A$ and $B$.
I know that tangent to parabola is $y = mx + a/m$ ($m$ being the slope), but I am not able to figure out how to take out point of intersections.
analytic-geometry conic-sections tangent-line
analytic-geometry conic-sections tangent-line
edited Dec 29 '18 at 23:30
Blue
47.7k870151
asked Dec 29 '18 at 15:56
BadguyBadguy
63
edited Dec 29 '18 at 23:30
Blue
47.7k870151
asked Dec 29 '18 at 15:56
BadguyBadguy
63
edited Dec 29 '18 at 23:30
Blue
47.7k870151
edited Dec 29 '18 at 23:30
Blue
47.7k870151
edited Dec 29 '18 at 23:30
Blue
47.7k870151
47.7k870151
asked Dec 29 '18 at 15:56
BadguyBadguy
63
asked Dec 29 '18 at 15:56
BadguyBadguy
63
asked Dec 29 '18 at 15:56
BadguyBadguy
63
63
1
$begingroup$
Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
$endgroup$
– Blue
Dec 29 '18 at 23:30
$begingroup$
I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
$endgroup$
– Badguy
Dec 30 '18 at 8:47
$begingroup$
What is "$T$" ?
$endgroup$
– Blue
Dec 30 '18 at 8:50
add a comment |
1
$begingroup$
Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
$endgroup$
– Blue
Dec 29 '18 at 23:30
$begingroup$
I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
$endgroup$
– Badguy
Dec 30 '18 at 8:47
$begingroup$
What is "$T$" ?
$endgroup$
– Blue
Dec 30 '18 at 8:50
1
1
$begingroup$
Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
$endgroup$
– Blue
Dec 29 '18 at 23:30
$begingroup$
Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
$endgroup$
– Blue
Dec 29 '18 at 23:30
$begingroup$
I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
$endgroup$
– Badguy
Dec 30 '18 at 8:47
$begingroup$
I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
$endgroup$
– Badguy
Dec 30 '18 at 8:47
$begingroup$
What is "$T$" ?
$endgroup$
– Blue
Dec 30 '18 at 8:50
$begingroup$
What is "$T$" ?
$endgroup$
– Blue
Dec 30 '18 at 8:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Given $$y^2=4ax$$ then $$2yy'=4a$$
answered Dec 29 '18 at 16:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
$endgroup$
$begingroup$
Not understood.
$endgroup$
– Badguy
Dec 29 '18 at 16:22
$begingroup$
I have neither studied nor seen what youve written here
$endgroup$
– Badguy
Dec 29 '18 at 16:23
$begingroup$
@Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
$endgroup$
– Ng Chung Tak
Jan 2 at 8:46
add a comment |
$begingroup$
$$y^2=4ax tag{1}$$
$$frac{x^2}{a^2}-frac{y^2}{b^2}=1 tag{2}$$
Let $P(X,Y)$ be the required locus.
For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).
Equation of $AB$ is
$$frac{X x}{a^2}-frac{Y y}{b^2}=1 tag{3}$$
Equation of tangent of $(1)$ at $C(x_1,y_1)$
$$y_1 y=2a(x+x_1)$$
Rearranging, we have
$$-frac{x}{x_1}+frac{y_1 y}{2a x_1}=1 tag{4}$$
Identifying $(3)$ and $(4)$, we get
$$(X,Y)=left( -frac{a^2}{x_1}, -frac{b^2 y_1}{2a x_1} right)$$
$$(x_1,y_1)=left( -frac{a^2}{X}, frac{2a^3 Y}{b^2 X} right)$$
But $$y_1^2=4a x_1$$
$$left( frac{2a^3 Y}{b^2 X} right)^2=4aleft( -frac{a^2}{X} right)$$
The locus of $P$ is
$$fbox{$a^3 Y^2+b^4 X=0$}$$
Useful fact:
Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by
$$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$
answered Jan 2 at 9:37
Ng Chung TakNg Chung Tak
14.3k31334
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
Hint: Given $$y^2=4ax$$ then $$2yy'=4a$$
answered Dec 29 '18 at 16:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
$endgroup$
$begingroup$
Not understood.
$endgroup$
– Badguy
Dec 29 '18 at 16:22
$begingroup$
I have neither studied nor seen what youve written here
$endgroup$
– Badguy
Dec 29 '18 at 16:23
$begingroup$
@Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
$endgroup$
– Ng Chung Tak
Jan 2 at 8:46
add a comment |
$begingroup$
Hint: Given $$y^2=4ax$$ then $$2yy'=4a$$
answered Dec 29 '18 at 16:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
$endgroup$
$begingroup$
Not understood.
$endgroup$
– Badguy
Dec 29 '18 at 16:22
$begingroup$
I have neither studied nor seen what youve written here
$endgroup$
– Badguy
Dec 29 '18 at 16:23
$begingroup$
@Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
$endgroup$
– Ng Chung Tak
Jan 2 at 8:46
add a comment |
$begingroup$
Hint: Given $$y^2=4ax$$ then $$2yy'=4a$$
answered Dec 29 '18 at 16:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
$endgroup$
Hint: Given $$y^2=4ax$$ then $$2yy'=4a$$
answered Dec 29 '18 at 16:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered Dec 29 '18 at 16:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered Dec 29 '18 at 16:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered Dec 29 '18 at 16:12
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
73.5k42865
$begingroup$
Not understood.
$endgroup$
– Badguy
Dec 29 '18 at 16:22
$begingroup$
I have neither studied nor seen what youve written here
$endgroup$
– Badguy
Dec 29 '18 at 16:23
$begingroup$
@Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
$endgroup$
– Ng Chung Tak
Jan 2 at 8:46
add a comment |
$begingroup$
Not understood.
$endgroup$
– Badguy
Dec 29 '18 at 16:22
$begingroup$
I have neither studied nor seen what youve written here
$endgroup$
– Badguy
Dec 29 '18 at 16:23
$begingroup$
@Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
$endgroup$
– Ng Chung Tak
Jan 2 at 8:46
$begingroup$
Not understood.
$endgroup$
– Badguy
Dec 29 '18 at 16:22
$begingroup$
Not understood.
$endgroup$
– Badguy
Dec 29 '18 at 16:22
$begingroup$
I have neither studied nor seen what youve written here
$endgroup$
– Badguy
Dec 29 '18 at 16:23
$begingroup$
I have neither studied nor seen what youve written here
$endgroup$
– Badguy
Dec 29 '18 at 16:23
$begingroup$
@Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
$endgroup$
– Ng Chung Tak
Jan 2 at 8:46
$begingroup$
@Badguy It's notation from (differential) calculus. $y'$ here is known as the derivative of $y(x)$ with respect to $x$, sometimes it's written as $dfrac{dy}{dx}$. $y'$ is equivalent to the slope at $(x,y)$ of the curve.
$endgroup$
– Ng Chung Tak
Jan 2 at 8:46
add a comment |
$begingroup$
$$y^2=4ax tag{1}$$
$$frac{x^2}{a^2}-frac{y^2}{b^2}=1 tag{2}$$
Let $P(X,Y)$ be the required locus.
For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).
Equation of $AB$ is
$$frac{X x}{a^2}-frac{Y y}{b^2}=1 tag{3}$$
Equation of tangent of $(1)$ at $C(x_1,y_1)$
$$y_1 y=2a(x+x_1)$$
Rearranging, we have
$$-frac{x}{x_1}+frac{y_1 y}{2a x_1}=1 tag{4}$$
Identifying $(3)$ and $(4)$, we get
$$(X,Y)=left( -frac{a^2}{x_1}, -frac{b^2 y_1}{2a x_1} right)$$
$$(x_1,y_1)=left( -frac{a^2}{X}, frac{2a^3 Y}{b^2 X} right)$$
But $$y_1^2=4a x_1$$
$$left( frac{2a^3 Y}{b^2 X} right)^2=4aleft( -frac{a^2}{X} right)$$
The locus of $P$ is
$$fbox{$a^3 Y^2+b^4 X=0$}$$
Useful fact:
Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by
$$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$
answered Jan 2 at 9:37
Ng Chung TakNg Chung Tak
14.3k31334
$endgroup$
add a comment |
$begingroup$
$$y^2=4ax tag{1}$$
$$frac{x^2}{a^2}-frac{y^2}{b^2}=1 tag{2}$$
Let $P(X,Y)$ be the required locus.
For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).
Equation of $AB$ is
$$frac{X x}{a^2}-frac{Y y}{b^2}=1 tag{3}$$
Equation of tangent of $(1)$ at $C(x_1,y_1)$
$$y_1 y=2a(x+x_1)$$
Rearranging, we have
$$-frac{x}{x_1}+frac{y_1 y}{2a x_1}=1 tag{4}$$
Identifying $(3)$ and $(4)$, we get
$$(X,Y)=left( -frac{a^2}{x_1}, -frac{b^2 y_1}{2a x_1} right)$$
$$(x_1,y_1)=left( -frac{a^2}{X}, frac{2a^3 Y}{b^2 X} right)$$
But $$y_1^2=4a x_1$$
$$left( frac{2a^3 Y}{b^2 X} right)^2=4aleft( -frac{a^2}{X} right)$$
The locus of $P$ is
$$fbox{$a^3 Y^2+b^4 X=0$}$$
Useful fact:
Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by
$$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$
answered Jan 2 at 9:37
Ng Chung TakNg Chung Tak
14.3k31334
$endgroup$
add a comment |
$begingroup$
$$y^2=4ax tag{1}$$
$$frac{x^2}{a^2}-frac{y^2}{b^2}=1 tag{2}$$
Let $P(X,Y)$ be the required locus.
For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).
Equation of $AB$ is
$$frac{X x}{a^2}-frac{Y y}{b^2}=1 tag{3}$$
Equation of tangent of $(1)$ at $C(x_1,y_1)$
$$y_1 y=2a(x+x_1)$$
Rearranging, we have
$$-frac{x}{x_1}+frac{y_1 y}{2a x_1}=1 tag{4}$$
Identifying $(3)$ and $(4)$, we get
$$(X,Y)=left( -frac{a^2}{x_1}, -frac{b^2 y_1}{2a x_1} right)$$
$$(x_1,y_1)=left( -frac{a^2}{X}, frac{2a^3 Y}{b^2 X} right)$$
But $$y_1^2=4a x_1$$
$$left( frac{2a^3 Y}{b^2 X} right)^2=4aleft( -frac{a^2}{X} right)$$
The locus of $P$ is
$$fbox{$a^3 Y^2+b^4 X=0$}$$
Useful fact:
Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by
$$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$
answered Jan 2 at 9:37
Ng Chung TakNg Chung Tak
14.3k31334
$endgroup$
$$y^2=4ax tag{1}$$
$$frac{x^2}{a^2}-frac{y^2}{b^2}=1 tag{2}$$
Let $P(X,Y)$ be the required locus.
For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).
Equation of $AB$ is
$$frac{X x}{a^2}-frac{Y y}{b^2}=1 tag{3}$$
Equation of tangent of $(1)$ at $C(x_1,y_1)$
$$y_1 y=2a(x+x_1)$$
Rearranging, we have
$$-frac{x}{x_1}+frac{y_1 y}{2a x_1}=1 tag{4}$$
Identifying $(3)$ and $(4)$, we get
$$(X,Y)=left( -frac{a^2}{x_1}, -frac{b^2 y_1}{2a x_1} right)$$
$$(x_1,y_1)=left( -frac{a^2}{X}, frac{2a^3 Y}{b^2 X} right)$$
But $$y_1^2=4a x_1$$
$$left( frac{2a^3 Y}{b^2 X} right)^2=4aleft( -frac{a^2}{X} right)$$
The locus of $P$ is
$$fbox{$a^3 Y^2+b^4 X=0$}$$
Useful fact:
Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x_1,y_1)$ is given by
$$ax_1 x+h(y_1 x+x_1 y)+by_1 y+g(x+x_1)+f(y+y_1)+c=0$$
answered Jan 2 at 9:37
Ng Chung TakNg Chung Tak
14.3k31334
edited Jan 2 at 9:47
answered Jan 2 at 9:37
Ng Chung TakNg Chung Tak
14.3k31334
answered Jan 2 at 9:37
Ng Chung TakNg Chung Tak
14.3k31334
answered Jan 2 at 9:37
Ng Chung TakNg Chung Tak
14.3k31334
14.3k31334
add a comment |
add a comment |
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Since you have a tangent line equation, you can find $A$ and $B$ by the standard substitution method: replace $y$ with $mx+a/m$ for $y$ in the hyperbola equation, and solve for $x$; then put the resulting $x$-values into the line equation to get the corresponding $y$-values. For the tangents at $A$ and $B$ ... There are many ways to find tangent line equations. Your comment to Dr. Graubner's answer indicates that you don't know calculus-based approaches. What approaches do you know? How did you get the formula for the tangent to the parabola? Help us help you.
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– Blue
Dec 29 '18 at 23:30
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I dont know the method of calculus but the basic method that tangent at any point on a curve is T=0.
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– Badguy
Dec 30 '18 at 8:47
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What is "$T$" ?
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– Blue
Dec 30 '18 at 8:50