JEE Advance - Mathematics (2005 - No. 13)
Find the equation of the common tangent in $${1^{st}}$$ quadrant to the circle $${x^2} + {y^2} = 16$$ and the ellipse $${{{x^2}} \over {25}} + {{{y^2}} \over 4} = 1$$. Also find the length of the intercept of the tangent between the coordinate axes.
$$y = -\frac{2}{\sqrt{3}}x + 4\sqrt{\frac{7}{3}}, \,\,\,\,\frac{14}{\sqrt{3}}$$
$$y = -\frac{1}{\sqrt{3}}x + 4\sqrt{\frac{7}{3}}, \,\,\,\,\frac{14}{\sqrt{3}}$$
$$y = -\frac{2}{\sqrt{3}}x + 2\sqrt{\frac{7}{3}}, \,\,\,\,\frac{7}{\sqrt{3}}$$
$$y = -\frac{2}{\sqrt{3}}x + 4\sqrt{\frac{7}{3}}, \,\,\,\,\frac{7}{\sqrt{3}}$$
$$y = -\frac{1}{\sqrt{3}}x + 2\sqrt{\frac{7}{3}}, \,\,\,\,\frac{14}{\sqrt{3}}$$
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