The acute angle between the pair of tangents drawn to the ellipse 2 x²+3 y²=5 from the point (1,3) is [26-Jul-2022-Shift-2]

Correct Answer is : tan⁡−1( 2475)^{-1} ≤ft( \; {24}{7 {5}} )tan−1(7524)

^{-1} ≤ft( \; {3+8 {5}}{35} )

Correct Answer is : tan⁡−1( 2475)^{-1} ≤ft( \; {24}{7 {5}} )tan−1(7524)

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Section: Mathematics

Question : 57 of 117

Marks: +1, -0

2 x^2+3 y^2=5

(1,3)

Solution: 👈: Video Solution

2 x^2+3 y^2=5

Equation of tangent having slope

m

y=m x \pm \sqrt{\; \frac{5}{2} m^2 + \; \frac{5}{3}}

which passes through

(1,3)

3=m \pm \sqrt{\; \frac{5}{2} m^2 + \; \frac{5}{3}}

\; \frac{5}{2} m^2 + \; \frac{5}{3} = 9 + m^2 - 6 m

\; \frac{3}{2} m^2 + 6 m - \; \frac{22}{3} = 0

9 m^2+36 m-44=0

m_1+m_2=-4, m_1 m_2=- \; \frac{44}{9}

(m_1-m_2)^2 = 16 + 4 \times \; \frac{44}{9} = \; \frac{320}{9}

Acute angle between the tangents is given by

\alpha = \tan^{-1} \left| \; \frac{m_1-m_2}{1+m_1 m_2} \right|

= \tan^{-1} \left| \; \frac{ \frac{8 \sqrt{5}}{3} }{ 1 - \; \frac{44}{9} } \right|

= \tan^{-1} \left( \; \frac{24 \sqrt{5}}{35} \right)

\alpha = \tan^{-1} \left( \; \frac{24}{7 \sqrt{5}} \right)

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