Question

The equation of tangent to the curve $9{x^2} + 16{y^2} = 144$ which makes equal intercepts with coordinate axes, is

Answer 1

Here we need to find the equation of the tangent to the given curve. Here first we will write the equation in the standard form. Then we will use the standard form of the tangent to a curve which includes the intercepts of the curve with the coordinate axes. Then we will equate both the intercepts, from there, we will get the value of all the constants and hence, the equation of the tangents.

Complete Step by Step Solution:
The equation of the given is $9{x^2} + 16{y^2} = 144$.
Now, we will write it in the standard form.
$\Rightarrow \dfrac{{{x^2}}}{{16}} + \dfrac{{{y^2}}}{9} = 1$
$\Rightarrow \dfrac{{{x^2}}}{{{4^2}}} + \dfrac{{{y^2}}}{{{3^2}}} = 1$…………. $\left( 1 \right)$

We know the standard form of equation of the ellipse is given by
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$………… $\left( 2 \right)$
On comparing equation $\left( 1 \right)$ with equation $\left( 2 \right)$, we get
$\begin{array}{l}a = 4\\\b = 3\end{array}$
We know that the standard equation of the tangent to ellipse is given by;
$\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta = 1$
Now, we will substitute the value of $a$ and $b$ here.
$\Rightarrow \dfrac{x}{4}\cos \theta + \dfrac{y}{3}\sin \theta = 1$ ………. $\left( 3 \right)$
On rearranging the terms, we get
$\Rightarrow \dfrac{x}{{\dfrac{4}{{\cos \theta }}}} + \dfrac{y}{{\dfrac{3}{{\sin \theta }}}} = 1$
Here, $\dfrac{4}{{\cos \theta }}$ and $\dfrac{3}{{\sin \theta }}$ are the intercepts of the curve to the coordinate axes.
As it is given that the given curve makes equal intercepts with the coordinate axes.
Therefore,
$\dfrac{4}{{\cos \theta }} = \dfrac{3}{{\sin \theta }}$
On cross multiplying the terms, we get
$\Rightarrow \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{3}{4}$ …….. $\left( 4 \right)$
We know the trigonometric formula $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$ .
Now, substituting $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$ in equation $\left( 1 \right)$, we get
$\Rightarrow \tan \theta = \dfrac{3}{4}$
Thus, the value of $\theta$ will be
$\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
Using the table of tangent of angels, we get
$\Rightarrow \theta = 37^\circ$
From the trigonometric table, we get
$\begin{array}{l}\sin 37^\circ = \dfrac{3}{5}\\\\\cos 37^\circ = \dfrac{4}{5}\end{array}$
On substituting these values in the equation $\left( 1 \right)$, we get
$\dfrac{x}{4} \times \dfrac{4}{5} + \dfrac{y}{3} \times \dfrac{3}{5} = 1$
On multiplying the terms, we get
$\Rightarrow \dfrac{x}{5} + \dfrac{y}{5} = 1$
On further simplification, we get
$\Rightarrow x + y = 5$

Hence, the equation of the tangent is equal to $x + y = 5$.

Note:
We have converted the given equation in the standard form of a tangent of an ellipse. In order to solve this question we need to remember both the general equation of the tangent and its parametric form. A tangent to a curve is defined as the straight line that touches the given curve at only one point. A tangent of a curve doesn’t cross the curve.