From point P(8, 27),tangents PQ and PR are dr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The equation of the chord of contact $QR$ for the point $P(8, 27)$ with respect to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ is given by

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The equation of the chord of contact $QR$ for the point $P(8, 27)$ with respect to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$ is given by $T = 0$: $\frac{8x}{4} + \frac{27y}{9} = 1 \Rightarrow 2x + 3y = 1$. The equation of the pair of lines passing through the origin and the points $Q$ and $R$ is obtained by homogenizing the ellipse equation using the chord of contact: $\frac{x^{2}}{4} + \frac{y^{2}}{9} = (2x + 3y)^{2}$. $9x^{2} + 4y^{2} = 36(4x^{2} + 12xy + 9y^{2})$. $9x^{2} + 4y^{2} = 144x^{2} + 432xy + 324y^{2}$. $135x^{2} + 432xy + 320y^{2} = 0$. Comparing this with $ax^{2} + 2hxy + by^{2} = 0$,we have $a = 135$,$2h = 432$ $(h = 216)$,and $b = 320$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^{2} - ab}}{a + b} ight|$. $	an 	heta = \left| \frac{2\sqrt{216^{2} - 135 \cdot 320}}{135 + 320} ight| = \left| \frac{2\sqrt{46656 - 43200}}{455} ight| = \frac{2\sqrt{3456}}{455} = \frac{2 \cdot 24 \sqrt{6}}{455} = \frac{48 \sqrt{6}}{455}$. Thus,$	heta = 	an^{-1} \left( \frac{48 \sqrt{6}}{455} ight)$,which is not among the given options.

From point $P(8, 27)$,tangents $PQ$ and $PR$ are drawn to the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$. The angle subtended by $QR$ at the origin is:

Similar Questions

An ellipse passing through $(4 \sqrt{2}, 2 \sqrt{6})$ has foci at $(-4, 0)$ and $(4, 0)$. Then,its eccentricity is

The equation $\frac{x^2}{2-r}+\frac{y^2}{r-5}+1=0$ represents an ellipse if

The eccentricity of the ellipse $4x^2 + 25y^2 = 100$ is

The equation of the chord of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$,whose mid-point is $(3, 1)$,is:

Let $S$ and $S^{\prime}$ be the foci of an ellipse and $B$ be one end of its minor axis. If $\triangle SBS^{\prime}$ is an isosceles right-angled triangle,then the eccentricity of the ellipse is

Vedclass Test Series

Exam Paper Generator

Online Exam Module