The equation of the tangent parallel to y-x+5 | vedclass

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

Question

(A) The given hyperbola is $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$. Here,$a^{2}=3$ and $b^{2}=2$. The equation of the tangent parallel to $y-x+5=0$ is of

Vedclass · Accepted Answer

$x-y-1=0$

Vedclass · Answer

(A) The given hyperbola is $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$. Here,$a^{2}=3$ and $b^{2}=2$. The equation of the tangent parallel to $y-x+5=0$ is of the form $y=x+c$. Comparing with $y=mx+c$,we have $m=1$. The condition for the line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}-b^{2}$. Substituting the values,$c^{2}=3(1)^{2}-2 = 3-2 = 1$. Thus,$c=\pm 1$. The equations of the tangents are $y=x+1$ or $y=x-1$,which can be written as $x-y+1=0$ or $x-y-1=0$. Comparing with the given options,$x-y-1=0$ is the correct choice.

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

Similar Questions

The area of the quadrilateral formed by the foci of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and its conjugate hyperbola is (in square units):

Area of the triangle formed by the lines $x - y = 0$,$x + y = 0$ and any tangent to the hyperbola $x^2 - y^2 = a^2$ is:

If the tangent at the point $(2 \sec \theta, 3 \tan \theta)$ to the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$ is parallel to $3x-y+4=0$,then the value of $\theta$ is (in $^{\circ}$)

$A$ hyperbola passes through a focus of the ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$. Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of their eccentricities is $1$. Then,the equation of the hyperbola is

The locus of a point of intersection of two lines $x \sqrt{3}-y=k \sqrt{3}$ and $\sqrt{3} k x+k y=\sqrt{3}$,where $k \in R$,describes:

Vedclass Test Series

Exam Paper Generator

Online Exam Module