Name: Conic Sections and Ellipse
Uploaded: 2019-07-19
Duration: 3 min 55 s
Description: In this video, we find the equation of an ellipse that is centred at the origin when the eccentricity and vertices are given. Study this video for a better understanding.

Question

Two lines are drawn at right angles, one being a tangent to

y^{2} = 4 a x

and the other to

x^{2} = 4 b y

. Show that the locus of their point of intersection is the curve

$(a x + b y) (x^{2} + y^{2}) + {(b x - a y)}^{2} = 0$

Accepted Answer

The equation of the tangent at $P (a {t_{1}}^{2}, 2 a t_{1})$ to the parabola $y^{2} = 4 a x$ is

$y t_{1} = x + a {t_{1}}^{2}$

$\Rightarrow x - y t_{1} + a {t_{1}}^{2} = 0$

$\Rightarrow a {t_{1}}^{2} - y t_{1} + x = 0 . . . . (i)$

Also, the equation of the tangent at $Q (2 b t_{2}, b {t_{2}}^{2})$ to the parabola $x^{2} = 4 a y$ is

$x t_{2} = y + b {t_{2}}^{2} . . . . (i i)$

It is given that the tangents $(i)$ and $(i i)$ are perpendicular, so their slopes will be equal to $- 1$ .

$\Rightarrow (\frac{1}{t_{1}}) . t_{2} = - 1$

$\Rightarrow t_{2} = - t_{1}$

Equation $(i i)$ reduces to

$- x t_{1} = y + b {t_{2}}^{2}$

$\Rightarrow x t_{1} + y + b {t_{2}}^{2} = 0$

$\Rightarrow b {t_{2}}^{2} + x t_{1} + y = 0 . . . . (i i i)$

Solving equations $(i)$ and $(i i)$ , we get

$\frac{{t_{1}}^{2}}{- (x^{2} + y^{2})} = \frac{t_{1}}{b x - a y} = \frac{1}{a x + b y}$

From first two equations, we get

$t_{1} = \frac{- (x^{2} + y^{2})}{b x - a y}$

And from last two equations, we get

$t_{1} = \frac{b x - a y}{a x + b y}$

Combining these two equations, we get

$\frac{- (x^{2} + y^{2})}{b x - a y} = \frac{b x - a y}{a x + b y}$

$\Rightarrow - (x^{2} + y^{2}) (a x + b y) = {(b x - a y)}^{2}$

$\Rightarrow (a x + b y) (x^{2} + y^{2}) + {(b x - a y)}^{2} = 0$ which is the required locus of the point of intersection of two tangents.

Two lines are drawn at right angles, one being a tangent to $y^{2} = 4 a x$ and the other to $x^{2} = 4 b y$ . Show that the locus of their point of intersection is the curve
$(a x + b y) (x^{2} + y^{2}) + {(b x - a y)}^{2} = 0$

Important Questions on Conic Section

Learn and Prepare for any exam you want !

Two lines are drawn at right angles, one being a tangent to y2=4ax and the other to x2=4by. Show that the locus of their point of intersection is the curve ax+byx2+y2+bx-ay2=0

Important Questions on Conic Section

Learn and Prepare for any exam you want !

Two lines are drawn at right angles, one being a tangent to $y^{2} = 4 a x$ and the other to $x^{2} = 4 b y$ . Show that the locus of their point of intersection is the curve
$(a x + b y) (x^{2} + y^{2}) + {(b x - a y)}^{2} = 0$