Which one of the following is the common tang | vedclass

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

Question

(B) The equation of a tangent to the ellipse $\frac{x^2}{a^2 + b^2} + \frac{y^2}{b^2} = 1$ is given by $y = mx \pm \sqrt{(a^2 + b^2)m^2 + b^2}$.If thi

Vedclass · Accepted Answer

$by = ax - \sqrt{a^4 + a^2b^2 + b^4}$

Vedclass · Answer

(B) The equation of a tangent to the ellipse $\frac{x^2}{a^2 + b^2} + \frac{y^2}{b^2} = 1$ is given by $y = mx \pm \sqrt{(a^2 + b^2)m^2 + b^2}$.If this line is also a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{a^2 + b^2} = 1$,then its condition of tangency $c^2 = A^2m^2 + B^2$ must be satisfied,where $A^2 = a^2$ and $B^2 = a^2 + b^2$.Thus,$(a^2 + b^2)m^2 + b^2 = a^2m^2 + (a^2 + b^2)$.Simplifying this,we get $a^2m^2 + b^2m^2 + b^2 = a^2m^2 + a^2 + b^2$,which implies $b^2m^2 = a^2$.Therefore,$m^2 = \frac{a^2}{b^2}$,so $m = \pm \frac{a}{b}$.Substituting $m = \pm \frac{a}{b}$ into the tangent equation:$y = \pm \frac{a}{b}x \pm \sqrt{(a^2 + b^2)\frac{a^2}{b^2} + b^2} = \pm \frac{a}{b}x \pm \sqrt{\frac{a^4 + a^2b^2 + b^4}{b^2}}$.Multiplying by $b$,we get $by = \pm ax \pm \sqrt{a^4 + a^2b^2 + b^4}$.Comparing this with the given options,$by = ax - \sqrt{a^4 + a^2b^2 + b^4}$ is a common tangent.

Which one of the following is the common tangent to the ellipses $\frac{x^2}{a^2 + b^2} + \frac{y^2}{b^2} = 1$ and $\frac{x^2}{a^2} + \frac{y^2}{a^2 + b^2} = 1$?

Similar Questions

With one focus of the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$ as the centre,a circle is drawn which is tangent to the hyperbola with no part of the circle being outside the hyperbola. The radius of the circle is

If the common tangents to the parabola $x^2 = 4y$ and the circle $x^2 + y^2 = 4$ intersect at the point $P$,then find the square of the slope of the line.

The triangle $PQR$ of area $A$ is inscribed in the parabola $y^2 = 4ax$ such that the vertex $P$ lies at the vertex of the parabola and the base $QR$ is a focal chord. The modulus of the difference of the ordinates of the points $Q$ and $R$ is:

If the points of intersection of two distinct conics $x^2+y^2=4b$ and $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ lie on the curve $y^2=3x^2$,then $3\sqrt{3}$ times the area of the rectangle formed by the intersection points is............................

Vedclass Test Series

Exam Paper Generator

Online Exam Module