The equation of the tangent parallel to y - x | vedclass

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

Question

(A) The given hyperbola is $\frac{x^2}{3} - \frac{y^2}{2} = 1$. Comparing this with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 3$ and $b^2

Vedclass · Accepted Answer

$x - y - 1 = 0$

Vedclass · Answer

(A) The given hyperbola is $\frac{x^2}{3} - \frac{y^2}{2} = 1$. Comparing this with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 3$ and $b^2 = 2$. The equation of the tangent parallel to $y - x + 5 = 0$ (which is $y = x - 5$,slope $m = 1$) is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$. Substituting $m = 1, a^2 = 3, b^2 = 2$: $y = 1 \cdot x \pm \sqrt{3(1)^2 - 2}$ $y = x \pm \sqrt{3 - 2}$ $y = x \pm 1$ This gives two equations: $x - y + 1 = 0$ and $x - y - 1 = 0$. Comparing with the given options,$x - y - 1 = 0$ is the correct answer.

The equation of the tangent parallel to $y - x + 5 = 0$ drawn to the hyperbola $\frac{x^2}{3} - \frac{y^2}{2} = 1$ is

Similar Questions

If $\frac{x^{2}}{36}-\frac{y^{2}}{k^{2}}=1$ is a hyperbola,then which of the following statements can be true?

Let the equation of two diameters of a circle $x^{2} + y^{2} - 2x + 2fy + 1 = 0$ be $2px - y = 1$ and $2x + py = 4p$. Then the slope $m \in (0, \infty)$ of the tangent to the hyperbola $3x^{2} - y^{2} = 3$ passing through the centre of the circle is equal to $......$

Let $S$ be the focus of the hyperbola $x^2 - 2y^2 = 1$ lying on the positive $X$-axis. Let $P(-1, 1)$ be a given point. Then the area of the triangle formed by the line $PS$ with the coordinate axes is (in sq. units)

Find the equation of the hyperbola satisfying the given conditions: Foci $(0, \pm \sqrt{10})$,passing through $(2, 3)$.

If $PN$ is the perpendicular from a point $P$ on the rectangular hyperbola $x^2 - y^2 = a^2$ to any of its asymptotes,then the locus of the midpoint of $PN$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module