The equation of the hyperbola whose directrix | vedclass

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

Question

(A) Let $P(x, y)$ be any point on the hyperbola. The focus is $S(1, 1)$ and the directrix is $2x + y - 1 = 0$.By the definition of a conic section,$SP

Vedclass · Accepted Answer

$7x^2 + 12xy - 2y^2 - 2x + 4y - 7 = 0$

Vedclass · Answer

(A) Let $P(x, y)$ be any point on the hyperbola. The focus is $S(1, 1)$ and the directrix is $2x + y - 1 = 0$.By the definition of a conic section,$SP = e \cdot PM$,where $PM$ is the perpendicular distance from $P$ to the directrix.$SP^2 = e^2 \cdot PM^2$$(x - 1)^2 + (y - 1)^2 = 3 \cdot \left( \frac{2x + y - 1}{\sqrt{2^2 + 1^2}} \right)^2$$(x^2 - 2x + 1 + y^2 - 2y + 1) = 3 \cdot \frac{(2x + y - 1)^2}{5}$$5(x^2 + y^2 - 2x - 2y + 2) = 3(4x^2 + y^2 + 1 + 4xy - 2y - 4x)$$5x^2 + 5y^2 - 10x - 10y + 10 = 12x^2 + 3y^2 + 3 + 12xy - 6y - 12x$Rearranging the terms to one side:$7x^2 + 12xy - 2y^2 - 2x + 4y - 7 = 0$

The equation of the hyperbola whose directrix is $2x + y = 1$,focus is $(1, 1)$,and eccentricity is $e = \sqrt{3}$ is:

Similar Questions

The equation of the normal to the curve $3x^2 - y^2 = 8$,which is parallel to the line $x + 3y = 10$,is

Let $a$ and $b$ respectively be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation $9e^2 - 18e + 5 = 0$. If $S(5, 0)$ is a focus and $5x = 9$ is the corresponding directrix of this hyperbola,then $a^2 - b^2$ is equal to

Find the locus of the point of intersection of the lines $\sqrt{3}x - y - 4\sqrt{3}k = 0$ and $\sqrt{3}kx + yk - 4\sqrt{3} = 0$ for different values of $k$.

If the straight line $x \cos \alpha + y \sin \alpha = p$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,then:

If the line $x-1=0$ is a directrix of the hyperbola $kx^{2}-y^{2}=6$,then the hyperbola passes through which of the following points?

Vedclass Test Series

Exam Paper Generator

Online Exam Module