Let the circle C touch the line x - y + 1 = 0 | vedclass

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

Question

(C) Let the centre of the circle $C$ be $(\alpha, 0)$ where $\alpha > 0$. Since the circle touches the line $x - y + 1 = 0$,the radius $r$ is the perp

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(C) Let the centre of the circle $C$ be $(\alpha, 0)$ where $\alpha > 0$. Since the circle touches the line $x - y + 1 = 0$,the radius $r$ is the perpendicular distance from $(\alpha, 0)$ to the line:$r = \left| \frac{\alpha - 0 + 1}{\sqrt{1^2 + (-1)^2}} \right| = \frac{\alpha + 1}{\sqrt{2}}$Thus,$r^2 = \frac{(\alpha + 1)^2}{2} \quad \dots(1)$The circle cuts a chord of length $L = \frac{4}{\sqrt{13}}$ along the line $3x - 2y + 1 = 0$. The perpendicular distance $d$ from $(\alpha, 0)$ to this line is:$d = \left| \frac{3\alpha - 2(0) + 1}{\sqrt{3^2 + (-2)^2}} \right| = \frac{3\alpha + 1}{\sqrt{13}}$Using the relation $r^2 = d^2 + (L/2)^2$:$r^2 = \frac{(3\alpha + 1)^2}{13} + \left( \frac{2}{\sqrt{13}} \right)^2 = \frac{(3\alpha + 1)^2 + 4}{13} \quad \dots(2)$Equating $(1)$ and $(2)$:$\frac{(\alpha + 1)^2}{2} = \frac{(3\alpha + 1)^2 + 4}{13}$$13(\alpha^2 + 2\alpha + 1) = 2(9\alpha^2 + 6\alpha + 1 + 4)$$13\alpha^2 + 26\alpha + 13 = 18\alpha^2 + 12\alpha + 10$$5\alpha^2 - 14\alpha - 3 = 0$$(5\alpha + 1)(\alpha - 3) = 0$Since $\alpha > 0$,we have $\alpha = 3$. Then $r^2 = \frac{(3 + 1)^2}{2} = 8$,so $r = 2\sqrt{2}$.For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the focus is $(\alpha, 0) = (3, 0)$,so $ae = 3$. The transverse axis length is $2a = 2r = 4\sqrt{2}$,so $a = 2\sqrt{2}$ and $a^2 = 8$.Since $a^2e^2 = 9$,we have $8e^2 = 9$,so $e^2 = \frac{9}{8}$.Using $b^2 = a^2(e^2 - 1) = 8(\frac{9}{8} - 1) = 8(\frac{1}{8}) = 1$.Thus,$2a^2 + 3b^2 = 2(8) + 3(1) = 16 + 3 = 19$.

Similar Questions

If the curves $\frac{x^2}{4}+\frac{y^2}{9}=1$ and $\frac{x^2}{16}-\frac{y^2}{k}=1$ cut each other orthogonally, then $k=$

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

The values of $\lambda$,for which the point $(\lambda, \lambda-2)$ lies inside the ellipse $4x^2+9y^2=36$ and outside the parabola $y^2=x$,satisfy:

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

$AB$ is a chord of the parabola $y^2 = 4ax$ with one endpoint $A$ at the vertex of the parabola. $BC$ is drawn perpendicular to $AB$ and meets the axis of the parabola at $C$. What is the projection of $BC$ on the axis of the parabola?

Vedclass Test Series

Exam Paper Generator

Online Exam Module