Question

The line $2x + \sqrt 6 y = 2$ is a tangent to the curve ${x^2} - 2{y^2} = 4$ . The point of contact is

1. $(4, - \sqrt 6 )$
2. $(7, - 2\sqrt 6 )$
3. $(2,3)$
4. $(\sqrt 6 ,1)$

Answer 1

Here, we are given an equation of line and also an equation of the tangent to the curve. We need to find the point of contact. First, from the given equation of line, we will convert the equation in one variable form and substituting this value in the equation of tangent to the curve. Lastly, we will again substitute this value in the first one. By doing so, we will get the values of x and y and thus, we will get the final output too.

Complete answer:
Given the equation of line is as below:
$2x + \sqrt 6 y = 2$
$\Rightarrow \sqrt 6 y = 2 - 2x$
$\Rightarrow y = \dfrac{{2 - 2x}}{{\sqrt 6 }}$ ---- (1)

Also, we are given that,
The equation of the tangent to the curve as below:
${x^2} - 2{y^2} = 4$

Substituting the value of y from equation (1), into the above equation, we will get,
$\Rightarrow {x^2} - 2{(\dfrac{{2 - 2x}}{{\sqrt 6 }})^2} = 4$
On evaluating this, we will get,
$\Rightarrow {x^2} - 2(\dfrac{{{{(2 - 2x)}^2}}}{{{{(\sqrt 6 )}^2}}}) = 4$
$\Rightarrow {x^2} - 2(\dfrac{{{{(2 - 2x)}^2}}}{6}) = 4$ $(\because 6 = \sqrt 6 \times \sqrt 6 )$
$\Rightarrow {x^2} - \dfrac{{{{(2 - 2x)}^2}}}{3} = 4$
Taking LCM as 3, we will write the equation as,
$\Rightarrow 3{x^2} - {(2 - 2x)^2} = 4(3)$
We know that, ${(a - b)^2} = {a^2} - 2ab + {b^2}$ and so applying this, we will get,
$\Rightarrow 3{x^2} - ({2^2} - 2(2)(2x) + {(2x)^2}) = 12$
$\Rightarrow 3{x^2} - (4 - 8x + 4{x^2}) = 12$
Removing the brackets, we will get,
$\Rightarrow 3{x^2} - 4 + 8x - 4{x^2} = 12$
By using transposition method, we will move the term from RHS to LHS, we will get,
$\Rightarrow 3{x^2} - 4 + 8x - 4{x^2} - 12 = 0$
On evaluating this equation, we will get,
$\Rightarrow - {x^2} + 8x - 16 = 0$
$\Rightarrow {x^2} - 8x + 16 = 0$
$\Rightarrow {(x - 4)^2} = 0$
$\Rightarrow x - 4 = 0$
$\Rightarrow x = 4$

Next, we will substitute the value of x in equation (1), we will get,
$\therefore y = \dfrac{{2 - 2(4)}}{{\sqrt 6 }}$
Removing the brackets, we will get,
$\Rightarrow y = \dfrac{{2 - 8}}{{\sqrt 6 }}$
On evaluating this equation, we will get,
$\Rightarrow y = \dfrac{{ - 6}}{{\sqrt 6 }}$
$\Rightarrow y = \dfrac{{( - 1)(\sqrt 6 \times \sqrt 6 )}}{{\sqrt 6 }}$ $(\because 6 = \sqrt 6 \times \sqrt 6 )$
$\Rightarrow y = - \sqrt 6$

Another Method:
Equation of tangent at $({x_1},{y_1})$ is $x{x_1} - 2y{y_1} = 4$

Comparing this, with the given equation
$2x + \sqrt 6 y = 2$
Multiply by 2 throughout, we will get,
$4x + 2\sqrt 6 y = 4$
Thus, we get ${x_1} = 4$ and ${y_1} = \sqrt 6$.
Hence, for the line $2x + \sqrt 6 y = 2$ and a tangent to the curve ${x^2} - 2{y^2} = 4$ , the point of contact is $(4, - \sqrt 6 )$ .
Thus, option 1 is the correct answer.

Note:
The tangent is considered only when it touches a curve at a single point or else it is said to be simply a line. The point where tangent meets the circle is called point of tangency. Tangent can be considered for any curved shape. Since tangent is a line, hence it also has its equation. There can be only one tangent at a point to a circle.