Question

A tangent to the ellipse ${x^2} + 4{y^2} = 4$ meets the ellipse ${x^2} + 2{y^2} = 6$ at P and Q the tangents at P and Q of the ellipse ${x^2} + 2{y^2} = 6$ is
A. ${90^ \circ }$
B. ${60^ \circ }$
C. ${45^ \circ }$
D. ${30^ \circ }$

Answer 1

Here in this question, we have to find the angle between where the two tangent of the ellipse at point P and Q, to solve this first we have to find the slope at the any point of the first ellipse by differentiating the equation of curve i.e., find $m = {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {x,y} \right)}}$ and later substitute a value of slope in a equation of tangent $y - {y_0} = m\left( {x - {x_0}} \right)$ where $\left( {{x_0},{y_0}} \right)$ is the point where tangent line passes through next find the point P and Q and later find the slope of tangent of second ellipse at point P and Q then by the product of these two slopes we can know the angle between them.

Complete step by step answer:
The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point.
Given,
consider a equation of ellipse-1
$\Rightarrow \,\,\,\;{x^2} + 4{y^2} = 4$-----(1)
Let us choose any point on this ellipse say $\left( {2,0} \right)$
Now, differentiate the equation (1) with respect to x, then
$\Rightarrow \,\,\,\;\dfrac{d}{{dx}}\left( {{x^2} + 4{y^2}} \right) = \dfrac{d}{{dx}}\left( 4 \right)$
$\Rightarrow \,\,\,\;\dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( {4{y^2}} \right) = \dfrac{d}{{dx}}\left( 4 \right)$
Differentiate using a formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ and $\dfrac{d}{{dx}}constant = 0$, then
$\Rightarrow \,\,\,\;2x + 8y\dfrac{{dy}}{{dx}} = 0$
Subtract both side by -2x, then
$\Rightarrow \,\,\,\;8y\dfrac{{dy}}{{dx}} = - 2x$
Divide both side by 8y
$\Rightarrow \,\,\,\;\dfrac{{dy}}{{dx}} = - \dfrac{{2x}}{{8y}}$
Now, the slope of the tangent at point $\left( {2,0} \right)$ is
$\Rightarrow \,\,\,\;m = {\dfrac{{dy}}{{dx}}_{\left( {2,0} \right)}} = - \dfrac{{2\left( 2 \right)}}{{8\left( 0 \right)}}$
$\Rightarrow \,\,\,\;m = {\dfrac{{dy}}{{dx}}_{\left( {2,0} \right)}} = - \dfrac{4}{0} = \infty$---------(2)
Now, the equation of the tangent at $\left( {2,0} \right)$ is having slope $- \dfrac{4}{0}$ is:
The equation of tangent is $y - {y_0} = m\left( {x - {x_0}} \right)$
Here ${y_0} = 0$, ${x_0} = 2$ and $m = - \dfrac{4}{0}$, on substituting we get
$\Rightarrow \,\,y - 0 = \left( { - \dfrac{4}{0}} \right)\left( {x - 2} \right)$
On cross multiplying, we have
$\Rightarrow \,\,0\left( {y - 0} \right) = - 4\left( {x - 2} \right)$
$\Rightarrow \,\, - 4\left( {x - 2} \right) = 0$
Divide both side by -4
$\Rightarrow \,\,x - 2 = 0$
Add both side by 2, then we have
$\Rightarrow \,\,x = 2$
Now, consider the equation of ellipse-2
$\Rightarrow \,\,\,\;{x^2} + 2{y^2} = 6$-------(3)
Substitute the x value in equation (3), we have
$\Rightarrow \,\,\,\;{\left( 2 \right)^2} + 2{y^2} = 6$
$\Rightarrow \,\,\,\;4 + 2{y^2} = 6$
Subtract both side by 4, then we have
$\Rightarrow \,\,\,\;2{y^2} = 6 - 4$
$\Rightarrow \,\,\,\;2{y^2} = 2$
Divide both side by 2
$\Rightarrow \,\,\,\;{y^2} = 1$
Take square root on both side, then
$\Rightarrow \,\,\,\;y = \pm \sqrt 1$
As we know, the 1 is a square number of itself.
$\Rightarrow \,\,\,\;y = \pm \,1$
Therefore, the point P and Q are $P\left( {2, - 1} \right)$ and $Q\left( {2,1} \right)$
Now, find the equation of slope at P and Q on the ellipse-2
Consider equation (3)
$\Rightarrow \,\,\,{x^2} + 2{y^2} = 6$
Differentiate the with respect to x
$\Rightarrow \,\,\,\dfrac{d}{{dx}}\left( {{x^2} + 2{y^2}} \right) = \dfrac{d}{{dx}}\left( 6 \right)$
$\Rightarrow \,\,\,\dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( {2{y^2}} \right) = \dfrac{d}{{dx}}\left( 6 \right)$
Differentiate using a formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ and $\dfrac{d}{{dx}}constant = 0$, then
$\Rightarrow \,\,\,2x + 4y\dfrac{{dy}}{{dx}} = 0$
Subtract 2x on both side, then
$\Rightarrow \,\,\,4y\dfrac{{dy}}{{dx}} = - 2x$
Divide both side by 4y
$\Rightarrow \,\,\,\dfrac{{dy}}{{dx}} = - \dfrac{{2x}}{{4y}}$
Now, the slope of the tangent at point $P\left( {2, - 1} \right)$ is
$\Rightarrow \,\,\,{m_1} = {\left( {\dfrac{{dy}}{{dx}}} \right)_{P\left( {2, - 1} \right)}} = - \dfrac{{2\left( 2 \right)}}{{4\left( { - 1} \right)}}$
$\Rightarrow \,\,\,{m_1} = {\left( {\dfrac{{dy}}{{dx}}} \right)_{P\left( {2, - 1} \right)}} = \dfrac{4}{4}$
$\Rightarrow \,\,\,{m_1} = {\left( {\dfrac{{dy}}{{dx}}} \right)_{P\left( {2, - 1} \right)}} = 1$-------(4)
the slope of the tangent at point $Q\left( {2,1} \right)$ is
$\Rightarrow \,\,\,{m_2} = {\left( {\dfrac{{dy}}{{dx}}} \right)_{Q\left( {2,1} \right)}} = - \dfrac{{2\left( 2 \right)}}{{4\left( 1 \right)}}$
$\Rightarrow \,\,\,{m_2} = {\left( {\dfrac{{dy}}{{dx}}} \right)_{Q\left( {2,1} \right)}} = - \dfrac{4}{4}$
$\Rightarrow \,\,\,{m_2} = {\left( {\dfrac{{dy}}{{dx}}} \right)_{Q\left( {2,1} \right)}} = - 1$--------(5)
since, multiply equation (4) and (5), then
$\Rightarrow \,\,\,{m_1} \times {m_2}$
$\Rightarrow \,\,\,1 \times - 1$
$\Rightarrow \,\,\, - 1$
If the product of their slope of the curve is -1, then the lines are non vertical perpendicular lines.
This shows that tangents at P and Q are perpendicular to each other, then the angle $\theta = {90^0}$

So, the correct answer is “Option A”.

Note: The concept of the equation of tangent comes under the concept of application of derivatives. Here the major part is differentiation must know the standard differentiation formulas and we should know about the general equation of a line tangent. And remember If two lines have the same slope, then the lines are non vertical parallel lines and If the slopes of two lines are opposite reciprocals of one another, or the product of their slopes is −1, then the lines are non vertical perpendicular lines.