Question

Locus of the mid points of the segments which are tangents to the ellipse $\frac{1}{2}x^2+y^2=1$ and which are intercepted between the coordinate axes is -

• A. $\frac{1}{2}x^2+\frac{1}{4}y^2=1$
• B. $\frac{1}{4}x^2-\frac{1}{2}y^2=1$
• C. $\frac{1}{3x^2}+\frac{1}{4y^2}=1$
• D. $\frac{1}{2x^2}+\frac{1}{4y^2}=1$

Answer 1

Answer: (D)

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

Answer 2

The equation of the given ellipse is $\frac{x^2}{2} + \frac{y^2}{1} = 1$. The equation of the tangent to the ellipse at a point $(x_0, y_0)$ on the ellipse is $\frac{xx_0}{2} + \frac{yy_0}{1} = 1$. To find the x-intercept, set $y=0$: $\frac{xx_0}{2} = 1 \implies x = \frac{2}{x_0}$. So, the x-intercept point is $P(\frac{2}{x_0}, 0)$. To find the y-intercept, set $x=0$: $yy_0 = 1 \implies y = \frac{1}{y_0}$. So, the y-intercept point is $Q(0, \frac{1}{y_0})$. The midpoint $M(h, k)$ of the segment $PQ$ is given by: $h = \frac{\frac{2}{x_0} + 0}{2} = \frac{1}{x_0}$ $k = \frac{0 + \frac{1}{y_0}}{2} = \frac{1}{2y_0}$ From these, we get $x_0 = \frac{1}{h}$ and $y_0 = \frac{1}{2k}$. Since $(x_0, y_0)$ lies on the ellipse, it must satisfy the ellipse equation: $\frac{x_0^2}{2} + y_0^2 = 1$ Substitute the expressions for $x_0$ and $y_0$: $\frac{(1/h)^2}{2} + \left(\frac{1}{2k}\right)^2 = 1$ $\frac{1}{2h^2} + \frac{1}{4k^2} = 1$ Replacing $(h, k)$ with $(x, y)$ to get the locus equation: $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$