Let a tangent drawn at any point on the ellip | vedclass

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

Question

(C) The equation of the given ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Let a point $P(5 \cos 	heta, 4 \sin 	heta)$ be on the ellipse. The equat

Vedclass · Accepted Answer

$(0,0)$

Vedclass · Answer

(C) The equation of the given ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$. Let a point $P(5 \cos 	heta, 4 \sin 	heta)$ be on the ellipse. The equation of the tangent to the ellipse at point $P$ is $\frac{x \cos 	heta}{5} + \frac{y \sin 	heta}{4} = 1$. Setting $y=0$ to find the $X$-axis intercept $Q$,we get $x = \frac{5}{\cos 	heta}$. Thus,$Q = (5 \sec 	heta, 0)$. Since $R$ is the image of $Q$ with respect to $y=x$,we swap the coordinates to get $R = (0, 5 \sec 	heta)$. The equation of a circle with diameter $QR$ is $(x - x_Q)(x - x_R) + (y - y_Q)(y - y_R) = 0$. Substituting the coordinates,we get $(x - 5 \sec 	heta)(x - 0) + (y - 0)(y - 5 \sec 	heta) = 0$. This simplifies to $x^2 + y^2 - (5 \sec 	heta)x - (5 \sec 	heta)y = 0$. For any value of $	heta$,the point $(0,0)$ satisfies this equation. Therefore,the circle always passes through the origin $(0,0)$.

Let a tangent drawn at any point on the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ cut the $X$-axis at $Q$. Let $R$ be the image of $Q$ with respect to $y=x$. If $S$ is a circle with $QR$ as its diameter,then the fixed point through which the circle $S$ passes is

Similar Questions

If a tangent having slope $m = \frac{1}{3}$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ is a normal to the circle $(x + 1)^2 + (y + 1)^2 = 1$,then $a^2$ lies in the interval:

If tangents are drawn to the ellipse $x^2+2y^2=2$,then the locus of the mid-points of the intercepts made by those tangents between the coordinate axes is

The tangent and normal to the ellipse $3x^2 + 5y^2 = 32$ at the point $P(2, 2)$ meet the $x$-axis at $Q$ and $R$,respectively. Then the area (in sq. units) of the triangle $PQR$ is

An ellipse is drawn by taking a diameter of the circle $(x - 1)^2 + y^2 = 1$ as its semi-minor axis and a diameter of the circle $x^2 + (y - 2)^2 = 4$ as its semi-major axis. If the center of the ellipse is at the origin and its axes are the coordinate axes,then the equation of the ellipse is:

The product of the perpendiculars from the two foci of the ellipse $\frac{x^2}{9} + \frac{y^2}{25} = 1$ on the tangent at any point on the ellipse is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module