Question

P and Q are two points on the ellipse $\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1$ such that sum of their ordinates is 3. Prove that the locus of the intersection of the tangents at P and Q is $9{x^2} + 25{y^2} = 150y$

Answer 1

Hint : A tangent is a straight line that touches a curve or curved surface at only one point. The ordinate is known as the y-coordinate of a point, we are given the sum of the ordinate of the points P and Q and we have to find the locus of the intersection of their tangents. So using the relation between the ordinates, we find out whether the given statement is true or not.

Complete step-by-step answer :
If $(h,k)$ is the point of intersection of the tangents at $\theta$ and $\phi$ , then
$\dfrac{h}{a} = \dfrac{{\cos (\dfrac{{\theta + \phi }}{2})}}{{\cos (\dfrac{{\theta - \phi }}{2})}}\,,\,\dfrac{k}{b} = \dfrac{{\sin (\dfrac{{\theta + \phi }}{2})}}{{\cos (\dfrac{{\theta - \phi }}{2})}} \\\ \Rightarrow {(\dfrac{h}{a})^2} = {(\dfrac{{\cos (\dfrac{{\theta + \phi }}{2})}}{{\cos (\dfrac{{\theta - \phi }}{2})}})^2}\,,\,{(\dfrac{k}{b})^2} = {(\dfrac{{\sin (\dfrac{{\theta + \phi }}{2})}}{{\cos (\dfrac{{\theta - \phi }}{2})}})^2} \\\$
On adding the above two equations, we get –
$\dfrac{{{h^2}}}{{{a^2}}} + \dfrac{{{k^2}}}{{{b^2}}} = \dfrac{{{{\cos }^2}(\dfrac{{\theta + \phi }}{2})}}{{{{\cos }^2}(\dfrac{{\theta - \phi }}{2})}} + \dfrac{{{{\sin }^2}(\dfrac{{\theta + \phi }}{2})}}{{{{\cos }^2}(\dfrac{{\theta - \phi }}{2})}} = \dfrac{{{{\cos }^2}(\dfrac{{\theta + \phi }}{2}) + {{\sin }^2}(\dfrac{{\theta + \phi }}{2})}}{{{{\cos }^2}(\dfrac{{\theta - \phi }}{2})}} \\\ \Rightarrow \dfrac{{{h^2}}}{{{a^2}}} + \dfrac{{{k^2}}}{{{b^2}}} = \dfrac{1}{{{{\cos }^2}(\dfrac{{\theta - \phi }}{2})}}...(1) \\\$
Now the sum of the ordinates is given as 3, that is –
$b\sin \theta + b\sin \phi = 3 \\\ b(\sin \theta + \sin \phi ) = 3 \;$
We are given that the equation of the ellipse is $\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1$ on comparing this equation with the general equation of an ellipse that is $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$ , we get $a = 5\,and\,b = 3$
So,
$3(\sin \theta + \sin \phi ) = 3 \\\ \Rightarrow \sin \theta + \sin \phi = 1 \;$
We know that, $\sin C + \sin D = 2\sin (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2})$
Using this relation in the above equation, we get –
$2\sin (\dfrac{{\theta + \phi }}{2})\cos (\dfrac{{\theta - \phi }}{2}) = 1 \\\ \Rightarrow \sin (\dfrac{{\theta + \phi }}{2}) = \dfrac{1}{{2\cos (\dfrac{{\theta - \phi }}{2})}} \;$
We know that,
$\dfrac{k}{b} = \dfrac{{\sin (\dfrac{{\theta + \phi }}{2})}}{{\cos (\dfrac{{\theta - \phi }}{2})}} \\\ \Rightarrow \dfrac{k}{b} = \dfrac{1}{{2{{\cos }^2}(\dfrac{{\theta - \phi }}{2})}} \\\ \Rightarrow \dfrac{1}{{{{\cos }^2}(\dfrac{{\theta - \phi }}{2})}} = \dfrac{{2k}}{b}...(2) \;$
Using (2) in (1) , we get –
$\dfrac{{{h^2}}}{{{a^2}}} + \dfrac{{{k^2}}}{{{b^2}}} = \dfrac{{2k}}{b}$
Thus, locus at the point $(h,k)$ is $\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = \dfrac{{2y}}{3}$
Or $\dfrac{{9{x^2} + 25{y^2}}}{{25 \times 9}} = \dfrac{{2y}}{3} \\\ 9{x^2} + 25{y^2} = \dfrac{{2 \times 25 \times 9 \times y}}{3} \\\ \Rightarrow 9{x^2} + 25{y^2} = 150y \;$
So, the correct answer is “ $9{x^2} + 25{y^2} = 150y$ ”.

Note : A plane curve surrounding two focal points is called an ellipse, it is usually oval-shaped. An ellipse is drawn by tracing a point moving in a plane such that the sum of its distance from the two focal points is constant. When the two focal points become the same, the ellipse is converted into a circle.