The locus of the middle point of the intercep | vedclass

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

Question

(C) The given ellipse is $x^2 + 2y^2 = 2$,which can be written as $\frac{x^2}{2} + \frac{y^2}{1} = 1$. Here $a^2 = 2$ and $b^2 = 1$.Let the point of c

Vedclass · Accepted Answer

$\frac{1}{2x^2} + \frac{1}{4y^2} = 1$

Vedclass · Answer

(C) The given ellipse is $x^2 + 2y^2 = 2$,which can be written as $\frac{x^2}{2} + \frac{y^2}{1} = 1$. Here $a^2 = 2$ and $b^2 = 1$.Let the point of contact be $R \equiv (\sqrt{2} \cos 	heta, \sin 	heta)$.The equation of the tangent at $R$ is $\frac{x \cos 	heta}{\sqrt{2}} + y \sin 	heta = 1$.The tangent intersects the $x$-axis at $A \equiv (\sqrt{2} \sec 	heta, 0)$ and the $y$-axis at $B \equiv (0, \csc 	heta)$.Let $Q(h, k)$ be the midpoint of $AB$. Then $h = \frac{\sqrt{2} \sec 	heta}{2} = \frac{\sec 	heta}{\sqrt{2}}$ and $k = \frac{\csc 	heta}{2}$.This implies $\cos 	heta = \frac{1}{h\sqrt{2}}$ and $\sin 	heta = \frac{1}{2k}$.Using $\cos^2 	heta + \sin^2 	heta = 1$,we get $\left(\frac{1}{h\sqrt{2}}ight)^2 + \left(\frac{1}{2k}ight)^2 = 1$.Thus,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

The locus of the middle point of the intercept of the tangents drawn to the ellipse $x^2 + 2y^2 = 2$ between the coordinate axes is:

Similar Questions

The locus of the midpoints of the chords of the hyperbola $x^2 - y^2 = a^2$ which touch the parabola $y^2 = 4ax$ is:

If $e$ and $e^{\prime}$ are the eccentricities of the ellipse $5x^2 + 9y^2 = 45$ and the hyperbola $5x^2 - 4y^2 = 45$ respectively,then $ee^{\prime}$ is equal to

Let $\lim_{x \to 2} \frac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5$ for some $r, p \in R$. If the set of all possible values of $q$, such that the roots of the equation $rx^2 - px + q = 0$ lie in $(0, 2)$, be the interval $(\alpha, \beta]$, then $4(\alpha + \beta)$ equals :

If the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$ and the hyperbola $\frac{x^2}{144} - \frac{y^2}{81} = \frac{1}{25}$ coincide,then the value of $b^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module