Question

Find equations of tangents drawn to the curve ${y^2}\, - \,2{x^2}\, - \,4y\, + \,8\, = \,0$ from the point $(\,1\,,\,2\,)$.

Answer 1

Hint: Differentiate the given equation and find the slope of the given equation. Also find the slope of tangent. Assume the point and find the value of the point from the given equation. By using the points that are given and obtained points, find the equation of tangents.

Useful Formula:
Basic differential formula is used for differentiating the equation. The slope of tangent formula $\dfrac{{y\, - \,{y_1}}}{{x\, - \,{x_1}}}$ to find the value of $m$.

Complete step by step solution:
Given that: The tangent drawn to the curve is ${y^2}\, - \,2{x^2}\, - \,4y\, + \,8\, = \,0$
The point which passes through the tangent curve is $(\,1\,,\,2\,)$

Let the tangent from $(\,1\,,\,2\,)$to the curve meet at the curve $(\,a\,,\,b\,)$
${b^2}\, - \,2{a^2}\, - 4b\, + \,8\, = 0\,\,\,\,\,\,\,\,\, \to (\,1\,)$
Slope of the tangent $(\,m\,)$to the curve at $(\,a\,,\,b\,)$is given by the value of $\dfrac{{dy}}{{dx}}$ at the point $(\,a\,,\,b\,)$.
Consider the given equation:
${y^2}\, - \,2{x^2}\, - \,4y\, + \,8\, = \,0$
Differentiate the above equation with respect to $x$, as follows:
$2y\dfrac{{dy}}{{dx}}\, - \,4x\, - \,4\dfrac{{dy}}{{dx}}\, = \,0$
Simplify the above equation by splitting the equation by $\dfrac{{dy}}{{dx}}$on left-hand side and the remaining parts in right hand side as follows:
$2y\,\dfrac{{dy}}{{dx}}\, - 4\dfrac{{dy}}{{dx}}\, = \,4x$

Take the $\dfrac{{dy}}{{dx}}$as common part from the left side:
$\dfrac{{dy}}{{dx}}\,(2y\, - \,4)\, = \,4x$
Now, get the $2y\, - \,4$from the left side to the right side to get the value of $\dfrac{{dy}}{{dx}}$:
$\dfrac{{dy}}{{dx}}\, = \,\dfrac{{4x}}{{2y\, - \,4}}$

We can rewrite the above equation as follows:
$m\, = \,\dfrac{{dy}}{{dx}}\, = \,\dfrac{{4x}}{{2y\, - \,4}}$
Now, modify the above equation in terms of $(\,a\,,\,b\,)$as follows:
At$(\,a\,,\,b\,)$, $m\, = \,\,\dfrac{{4a}}{{2b\, - \,4}}\,\,\,\,\,\,\,\,\,\,\, \to (\,2\,)$
Since the tangents passes through the points $(\,1\,,\,2\,)$and $(\,a\,,\,b\,)$. Consider these two points that pass through the tangent curve.
Find the slope of tangent using the formula:
$m\, = \,\dfrac{{y\, - \,{y_1}}}{{x\, - \,{x_1}}}$
Now substitute the two points in the above formula to find the slope of tangent value:
$m\, = \,\dfrac{{b\, - \,2}}{{a\, - \,1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to (\,3\,)$
From the equation $(\,2\,)$and $(\,3\,)$, calculate the slope value:
$\dfrac{{4a}}{{2b\, - \,4}}\, = \,\dfrac{{b\, - \,2}}{{a\, - \,1}}$

Multiply the $(\,a\, - \,1\,)$in left side and multiply the $(\,2b\, - \,4\,)$in right side as follows:
$4a(a\, - \,1\,)\, = \,(b\, - \,2\,)(2b\, - \,4)$
Now, simplify the above equation by multiplying the terms involved in the equation
$4{a^2}\, - \,4a\, = \,2{b^2} - \,4b\, - \,4b\, + \,8$
Modify the equation by adding the terms:
$4{a^2}\, - \,4a\, = \,2{b^2}\, - \,8b\, + \,8$
Divide the above equation by$(\,2\,)$:
$2{a^2}\, - \,2a\, = \,{b^2}\, - \,4b\, + \,4$
Get all the terms in left side as follows:
${b^2}\, - \,2{a^2} - \,4b\, + \,2a\, + \,4\, = \,0\,\,\,\,\,\,\,\,\,\,\,\,\, \to (\,4\,)$

Subtract the equation $(\,1\,)$from $(\,4\,)$to get the value of $a$
$
  \,\,\,\,\,\,\,{b^2}\, - \,2{a^2} - \,4b\, + \,2a\, + \,4\, = \,0\, \\
  ( - )\,{b^2}\, - \,2{a^2}\, - 4b\, + \,0a + \,8\, = 0\, \\
 $
Cancel the ${b^2}\,,\,2{a^2}\,,\,4b$from the equation, it will become as follows;
$2a\, - \,4\, = \,0$

Assign the number on right side:
$2a\, = \,4$
Divide the $4$by$2$, to get the value of $a$
$a\, = \,2$
Substitute the value of $a\, = \,2$in equation$(\,1\,)$:
${b^2}\, - \,2{a^2}\, - 4b\, + \,8\, = 0$

Add the $a$ value to the above equation:
${b^2}\, - \,2({2^2})\, - \,4b\, + \,8\, = \,0$
Simplify the above equation as possible:
${b^2}\, - \,2(4)\, - \,4b\, + \,8\, = \,0$
Modify the above equation:
${b^2}\, - \,8\, - \,4b\, + \,8\, = \,0$

Cancel $8$ from the equation:
${b^2}\, - \,4b\, = \,0$
Take $b$as common from the equation:
$b(b\, - \,4)\, = \,0$
The above equation as the two values in it:
$b\, = \,0$

This is one value of $b$
$b\, - \,4\, = \,0$
Get the number to right side of equation;
$b\, = \,4$
Thus, the value of $b$is $4$or $0$
There are two possible tangents from $(\,1\,,\,2\,)$to the curve:
Thus, the points we obtained by solving the equation is $(\,2\,,\,4\,)$and $(\,2\,,\,0\,)$
Equation of the two tangents can be found out using the equation of a line passing through two points.
Tangents passing through two points $(\,1\,,\,2\,)\,(\,2\,,\,4\,)$
$\dfrac{{y\, - \,2}}{{x\, - \,1}}\, = \,\dfrac{{4\, - \,2}}{{2\, - \,1}}$
Simplify the above equation:
$\dfrac{{y\, - \,2}}{{x\, - \,1}}\, = \,2$
Get the above equation is in the form of tangents equation:
$y\, - \,2\, = \,2x\, - \,2$
Get all the terms in one side to get full form of equation:
$2x\, - \,y\, = 0$
(or)
$y\, = \,2x$

Tangents passing through two points $(\,1\,,\,2\,)\,(\,2\,,\,0\,)$
$\dfrac{{y\, - \,2}}{{x\, - \,1}}\, = \,\dfrac{{0\, - \,2}}{{2\, - \,1}}$
Simplify the above equation as possible:
$\dfrac{{y\, - \,2}}{{x\, - \,1}}\, = \, - 2$
Get the above equation is in the form of tangents equation:
$y\, - \,2\, = \,2\, - \,2x$
Get all the terms in one side to get full form of equation:
$y\, - \,2x\, - \,4\, = \,0$

Thus, there are two possible tangents from $(\,1\,,\,2\,)$ to the given curve and the equations are $y\, - \,2x\, - \,4\, = \,0$ and $y\, = \,2x$.

Note: By subtracting the two equations, the sign of all the terms in one of the equations should be changed to their opposite sign. If the points contain more than one value in it, we need to consider both the two values as the points.