Question

Find the equation of the tangent to the hyperbola $4{x^2} - 9{y^2} = 1$, which is parallel to the line $4y = 5x + 7$

Answer 1

Here the keyword was Hyperbola, is a conic section in which the difference of distances of all points from two fixed points called ‘foci’ is constant.
First we have to find the line equation to given equation for getting the value for $m$
Also, we have to find the equation of the hyperbola compared to the given equation.
On some simplification we get the equation of the tangent.

Formula used: A line is of the form $y = mx + c$ which is parallel to the hyperbola.
General equation of hyperbola is ${\dfrac{x}{{{a^2}}}^2} - \dfrac{{{y^2}}}{{{b^2}}} = 1$
The equation of tangent $y = mx \pm \sqrt {{a^2}{m^2} - {b^2}}$

Complete step-by-step answer:
It is given the line equation is
$4y = 5x + 7$
Divide the equation by $4$
$y = \dfrac{5}{4}x + \dfrac{7}{4}$
It is of the form, the line equation $y = mx + c$,
So we compared this and write as $m = \dfrac{5}{4}$ and $c = \dfrac{7}{4}$
Now, we have to find ${a^2},{b^2}$
So we take the given hyperbolic equation
$4{x^2} - 9{y^2} = 1$
The numerator of the number is the fraction of the denominator to the number, that is
${\dfrac{x}{{\dfrac{1}{4}}}^2} - \dfrac{{{y^2}}}{{\dfrac{1}{9}}} = 1....\left( 1 \right)$
Here the general equation of hyperbola: ${\dfrac{x}{{{a^2}}}^2} - \dfrac{{{y^2}}}{{{b^2}}} = 1$
Compare general equation of the hyperbola to the equation $\left( 1 \right)$
We can write it as,
${a^2} = \dfrac{1}{4}$,${b^2} = \dfrac{1}{9}$
Now, we have to find the equation of tangent
$y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} ....\left( 2 \right)$
Here the values are, $m = \dfrac{5}{4}$, ${a^2} = \dfrac{1}{4}$ and ${b^2} = \dfrac{1}{9}$
Substitute it in the equation $\left( 2 \right)$
$y = \dfrac{5}{4}x \pm \sqrt {\dfrac{1}{4}\left( {\dfrac{{{5^2}}}{{{4^2}}}} \right) - \dfrac{1}{9}}$
Squaring the bracket terms we get,
$y = \dfrac{5}{4}x \pm \sqrt {\dfrac{1}{4}\left( {\dfrac{{25}}{{16}}} \right) - \dfrac{1}{9}}$
Om multiplying the open brackets, we get
$y = \dfrac{5}{4}x \pm \sqrt {\dfrac{{25}}{{64}} - \dfrac{1}{9}}$
Taking L.C.M on the square root terms, we get
$y = \dfrac{5}{4}x \pm \sqrt {\dfrac{{25 \times 9}}{{64 \times 9}} - \dfrac{{64}}{{64 \times 9}}}$
On multiplying the numerator terms,
$y = \dfrac{5}{4}x \pm \sqrt {\dfrac{{225 - 64}}{{576}}}$
Let us subtract the numerator terms,
$y = \dfrac{5}{4}x \pm \sqrt {\dfrac{{161}}{{576}}}$
Split the square root terms,
$y = \dfrac{5}{4}x \pm \dfrac{{\sqrt {161} }}{{\sqrt {576} }}$
Here, we can write$\sqrt {576} = 24$
$y = \dfrac{5}{4}x \pm \dfrac{{\sqrt {161} }}{{24}}$
Again taking L.C.M. we get,
$y = \dfrac{{5 \times 6}}{{4 \times 6}}x \pm \dfrac{{\sqrt {161} }}{{24}}$
Multiplying the terms we get,
$y = \dfrac{{30x \pm \sqrt {161} }}{{24}}$
Let us take a cross multiply we can write it as,
$24y = 30x \pm \sqrt {161}$
Taking variable to the LHS,
$24y - 30x = \pm \sqrt {161}$
(Or)
$30x - 24y = \mp \sqrt {161}$
Thus, equation of tangent is $30x - 24y = \mp \sqrt {161}$ (or) $24y - 30x = \pm \sqrt {161}$

Note: A tangent line is a line which locally touches a curve at one and only one point.
Here, Tangent is a line which touches the hyperbola at four particular point
Here, the equation has $24y - 30x = \pm \sqrt {161}$ can also be written has $24y - 30x = + \sqrt {161}$,$24y - 30x = - \sqrt {161}$
And the equation has $30x - 24y = \mp \sqrt {161}$ and it can also be written has $30x - 24y = + \sqrt {161}$,$30x - 24y = - \sqrt {161}$
These $4$ points are the equation of the tangent which touches the hyperbola.