Question

From a point on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, two tangents are drawn to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the corresponding chord of contact meets $\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$ at points Q and R then

• A. The locus of mid-point of QR is $x^2 + y^2 = a^2$
• B. The locus of mid-point of QR is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
• C. The locus of image of mid-point of QR in x-axis is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
• D. The locus of foot of perpendicular from point $(\sqrt{a^2 + b^2},0)$ to QR is $x^2 + y^2 = a^2$

Answer 1

Answer: (B), (C)

The correct answers are (B) and (C).

Answer 2

Let $P(x_0, y_0)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. The chord of contact from $P$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1$. This chord of contact intersects the pair of lines $\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$ at points Q and R. Let $M(h, k)$ be the mid-point of QR. The equation of the chord of the pair of lines $\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$ with mid-point $(h, k)$ is $\frac{xh}{a^2} - \frac{yk}{b^2} = \frac{h^2}{a^2} - \frac{k^2}{b^2}$. Comparing this with the equation of the chord of contact $\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1$, we get $\frac{h}{x_0} = \frac{-k}{y_0} = \frac{h^2/a^2 - k^2/b^2}{1}$. Since $(x_0, y_0)$ is on the hyperbola, $\frac{x_0^2}{a^2}-\frac{y_0^2}{b^2}=1$. Substituting $x_0 = \frac{h}{h^2/a^2 - k^2/b^2}$ and $y_0 = \frac{-k}{h^2/a^2 - k^2/b^2}$ gives $\frac{1}{a^2}(\frac{h}{h^2/a^2 - k^2/b^2})^2 - \frac{1}{b^2}(\frac{-k}{h^2/a^2 - k^2/b^2})^2 = 1$, which simplifies to $\frac{h^2}{a^2} - \frac{k^2}{b^2} = 1$. Thus, the locus of the mid-point of QR is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

For option (C), the image of $(h, k)$ in the x-axis is $(h, -k)$. The locus of the image is found by substituting $x=h$ and $y=-k$ in the locus of $(h, k)$, which is $\frac{x^2}{a^2}-\frac{(-y)^2}{b^2}=1$, or $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.