Question

10. If two points \( P \& Q \) on the hyperbola \( \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \) whose centre is \( C \) be such that \( C P \) is perpendicular to \( CQ \& a < b \), then prove that \( \frac{1}{ CP ^{2}}+\frac{1}{ CQ ^{2}}=\frac{1}{ a ^{2}}-\frac{1}{ b ^{2}} \).

11. An ellipse has eccentricity \( 1 / 2 \) and one focus at the point \( P(1 / 2,1) \). Its one directrix is the common tangent, nearer to the point \( P \), to the circle \( x^{2}+y^{2}=1 \) and the hyperbola \( x^{2}-y^{2}=1 \). Find the equation of the ellipse in the standard form.

Answer 1

10. Given: \(\frac {x^2}{a^2} - \frac{y^2}{b^2} = 1\)

Let CP = r₁ be inclined to transverse axis at an angle θ.

P = (r₁cosθ, r₁sinθ) and P lies on the hyperbola, so

\(r_1^2 \left(\frac{\cos^2 \theta}{a^2} - \frac{\sin ^2 \theta}{b^2}\right) = 1\)

Replacing θ by 90° + θ, we get

Q = (−r₂sinθ, r₂cosθ) and Q also lies on the hyperbola, so

\(r_2^2 \left(\frac{\sin^2 \theta}{a^2} - \frac{\cos^2 \theta}{b^2}\right) = 1\)

Now,

11.

These are two common tangents to the circle x² + y² = 1 and the hyperbola x² − y² = 1.

There are x = 1 and x = −1

Out of these, x = 1 is nearer to the point P(\(\frac 12\), 1).

Thus, a directrix of the required ellipse is x = 1.

If Q(x, y) is any point on the ellipse, then its distance from the focus is QP = \(\sqrt{(x - \frac 12)^2 + (y - 1)^2}\) and its distance from the directrix x = 1 is |x − 1|

By definition of ellipse, QP = e|x − 1|