Question

A tangent to the ellipse $\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{4} = 1$ with the center C meets its director circle at P and Q. Then the product of the slopes of CP and CQ is:
A.$\dfrac{9}{4}$
B.$\dfrac{{ - 4}}{9}$
C.$\dfrac{2}{9}$
D.$\dfrac{{ - 1}}{4}$

Answer 1

Hint: Tangent is the line which touches the ellipse at one point. The locus of the point of intersection of two perpendicular tangents to a circle is called the Director circle. The slope of a line is a number that describes both the direction and the steepness of the line. Slope is calculated by finding the ratio of the "vertical change" to the "horizontal change" between (any) two distinct points on a line. The standard form of ellipse is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ and the standard form of director of circle is ${x^2} + {y^2} = {a^2} + {b^2}$ .

Complete answer:Standard form of ellipse $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
Assume two points on the ellipse $(a\cos \theta ,b\sin \theta )$
The equation of tangent will be
T=0
$\dfrac{{{x^2}}}{a}\cos \theta + \dfrac{{{y^2}}}{b}\sin \theta = 1$
Equation of director of circle
${x^2} + {y^{^2}} = {a^2} + {b^2}$
Given equation of ellipse is
$\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{4} = 1$
By comparing. we get,
Where $a = 3,b = 2$
$\dfrac{{{x^2}}}{3}\cos \theta + \dfrac{{{y^2}}}{2}\sin \theta = 1......(i)$
${x^2} + {y^2} = 13......(ii)$
We can write equation (ii) as
${x^2} + {y^2} = 13{(1)^2}$
Substituting the value of 1 from equation (i)
${x^2} + {y^2} = 13{\left( {\dfrac{x}{3}\cos \theta + \dfrac{y}{2}\sin \theta } \right)^2}$
Using the identity of ${(a + b)^2} = {a^2} + {b^2} + 2ab$
\[{x^2} + {y^2} = 13\left( {\dfrac{{{x^2}}}{9}{{\cos }^2}\theta + \dfrac{{{y^2}}}{4}{{\sin }^2}\theta + \dfrac{{xy}}{3}\cos \theta \sin \theta } \right)\]
Solving it further. We get,
 ${x^2}\left( {\dfrac{{13}}{9}{{\cos }^2}\theta - 1} \right) + {y^2}\left( {\dfrac{{13}}{4}{{\sin }^2}\theta - 1} \right) + \dfrac{{13xy}}{3}\cos \theta \sin \theta = 0$
To find the product of CP and CQ is

$\dfrac{{\text{coefficient of }{x^2}}}{{\text{coefficient of }{y^2}}} = \dfrac{{\dfrac{{13}}{9}{{\cos }^2}\theta - 1}}{{\dfrac{{13}}{4}{{\sin }^2}\theta - 1}}$
$ = \dfrac{4}{9}\left[ {\dfrac{{13{{\cos }^2}\theta - 9}}{{13{{\sin }^2}\theta - 4}}} \right]$
Using the identity of ${\cos ^2}\theta = 1 - {\sin ^2}\theta $
$ = \dfrac{4}{9}\left[ {\dfrac{{13(1 - {{\sin }^2}\theta ) - 9}}{{13{{\sin }^2}\theta - 4}}} \right]$
Multiplying 13 in the bracket
$ = \dfrac{4}{9}\left[ {\dfrac{{13 - 13{{\sin }^2}\theta - 9}}{{13{{\sin }^2}\theta - 4}}} \right]$
Solving it further,
$ = \dfrac{4}{9}\left[ {\dfrac{{ - 13{{\sin }^2}\theta + 4}}{{13{{\sin }^2}\theta - 4}}} \right]$
Divide numerator by denominator.
$ = \dfrac{{ - 4}}{9}$
The product of slope of CP and CQ is $\dfrac{{ - 4}}{9}$.

Note:
Ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the focal points is a constant. As such, it generalizes a circle, which is the special type of ellipse in which the two focal points are the same. The elongation of an ellipse is measured by its eccentricity $e$. Ellipses are the closed type of conic section: a plane curve tracing the intersection of a cone with a plane. Ellipses have many similarities with the other two forms of conic sections, parabolas and hyperbolas, both of which are open and unbounded. An angled cross section of a cylinder is also an ellipse.