QJEEM18Mac05d: JEE Main 2018 Mathematics - Class 11 JEEM18Mac05d Solution: If the tangents drawn to the hyperbola 4y 2 = x 2 + 1 intersect …

Equation of hyperbola is :

4y² = x² + 1 $\Rightarrow$ $-$ x² + 4y² = 1

$\Rightarrow$ $\,\,\,$ $-$ ${{{x^2}} \over {{1^2}}}$ + ${{{y^2}} \over {{{\left( {{1 \over 2}} \right)}^2}}}$ = 1

$\therefore$   a = 1, b = ${1 \over 2}$

Now, tangent to the curve at point (x₁, y₁) is given by

4 $\times$ 2y₁ ${{dy} \over {dx}}$ = 2x₁

$\Rightarrow$    ${{dy} \over {dx}}$ = ${{2{x_1}} \over {8{y_1}}}$ = ${{{x_1}} \over {4{y_1}}}$

Equation of tangent at (x₁, y₁) is

y = mx + c

$\Rightarrow$    y = ${{{x_1}} \over {4{y_1}}}$ . x + c

As tangent passes through (x₁, y₁)

$\therefore$    y₁ = ${{{x_1}{x_1}} \over {4{y_1}}} + c$

$\Rightarrow$    C = ${{4y_1^2 - x_1^2} \over {4{y_1}}}$ = ${1 \over {4{y_1}}}$

Therefore, y = ${{{x_1}} \over {4{y_1}}}x + {1 \over {4{y_1}}}$

$\Rightarrow$  4y₁y = x₁x + 1

which intersects x axis at A $\left( {{{ - 1} \over {{x_1}}},0} \right)$ and y axis at $B\left( {0,{1 \over {4{y_1}}}} \right)$

Let midpoint of AB is (h, k)

$\therefore$    h = ${{ - 1} \over {2{x_1}}}$

$\Rightarrow$ $\,\,\,$ x₁ = ${{ - 1} \over {2h}}$ & y₁ = ${1 \over {8k}}$

Thus, 4 ${\left( {{1 \over {8k}}} \right)^2}$ = ${\left( {{{ - 1} \over {2h}}} \right)^2}$ + 1

$\Rightarrow$ $\,\,\,$ ${1 \over {16{k^2}}}$ = ${1 \over {4{h^2}}}$ + 1

$\Rightarrow$ $\,\,\,$ 1 = ${{16{k^2}} \over {4{h^2}}}$ + 16k²

$\Rightarrow$ $\,\,\,$ h² = 4k² + 16h² k.

So, required equation is

x² $-$ 4y² $-$ 16x² y² = 0