If y = x + c touches the ellipse 3x^ 2 + 4y^ 2 = 12 at the point P,...

If $y = x + c$ touches $\dfrac{x^{2}}{4} + \dfrac{y^{2}}{3} = 1$

then,~$c = \pm \sqrt{4\left(1\right) + 3}$

$= \pm \sqrt{7}$

Let's assume,~$c = \sqrt{7}$~ and $y = x + \sqrt{7}$ touches $\dfrac{x^{2}}{4} + \dfrac{y^{2}}{3} = 1$ at \left(h, k\right) .$<br /><br />then,~$\dfrac{xh}{4} + \dfrac{yk}{3} = 1$is same as$y - x = \sqrt{7}$<br /><br />$⇒ \dfrac{\dfrac{h}{4}}{- 1} = \dfrac{\dfrac{k}{3}}{1} = \dfrac{1}{\sqrt{7}}$<br /><br />$⇒ \left(h, k\right) = \left(- \dfrac{4}{\sqrt{7}}, \dfrac{3}{\sqrt{7}}\right) = P$<br /><br />Hence, OP = \sqrt{\dfrac{16}{7} + \dfrac{9}{7}} = \dfrac{5}{\sqrt{7}}$& units