If the line a x + 4 y = 7 , where a touches the ellipse 3 x 2 + 4 y 2 = 1 at the point P in the first quadrant, then one of the focal distances of P is : [2026]

Question

If the line $a x + 4 y = \sqrt{7}$ , where $a \in R$ touches the ellipse $3 x^{2} + 4 y^{2} = 1$ at the point P in the first quadrant, then one of the focal distances of P is : [2026]

Smart Achievers · Accepted Answer

(1)

$α x + 4 y - \sqrt{7} = 0 touches 3 x^{2} + 4 y^{2} = 1$

$∴$ $c^{2} = a^{2} m^{2} + b^{2}$

$\frac{7}{16} = \frac{1}{3} \times \frac{α^{2}}{16} + \frac{1}{4} \Rightarrow α = 3, - 3$

$Tangent is 3 x + 4 y - \sqrt{7} = 0$

Let the point of contact be $P (x_{1}, y_{1})$

$∴$ $Tangent is 3 x x_{1} + 4 y y_{1} = 1$

$∴$ $\frac{3 x_{1}}{3} = \frac{4 y_{1}}{4} = \frac{1}{\sqrt{7}}$ $∴$ $P (\frac{1}{\sqrt{7}}, \frac{1}{\sqrt{7}})$

$e = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}$

$P S = e (P M)$

$= e (\frac{a}{e} - \frac{1}{\sqrt{7}})$

$= \frac{1}{2} (\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{7}}) = \frac{1}{\sqrt{3}} - \frac{1}{2 \sqrt{7}}$

$P S' = e (P M')$

$= \frac{1}{2} (\frac{a}{e} + \frac{1}{\sqrt{7}}) = \frac{1}{2} (\frac{1}{\sqrt{7}} + \frac{2}{\sqrt{3}})$

$= \frac{1}{\sqrt{3}} + \frac{1}{2 \sqrt{7}}$

Solution

Q.
If the line $a x + 4 y = \sqrt{7}$ , where $a \in R$ touches the ellipse $3 x^{2} + 4 y^{2} = 1$ at the point P in the first quadrant, then one of the focal distances of P is : [2026]

1 $\frac{1}{\sqrt{3}} + \frac{1}{2 \sqrt{7}}$

2 $\frac{1}{\sqrt{3}} + \frac{1}{2 \sqrt{5}}$

3 $\frac{1}{\sqrt{3}} - \frac{1}{2 \sqrt{11}}$

4 $\frac{1}{\sqrt{3}} - \frac{1}{2 \sqrt{5}}$

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Solution

Q. If the line ax+4y=7, where a∈R touches the ellipse 3x2+4y2=1 at the point P in the first quadrant, then one of the focal distances of P is : [2026]

1 13+127

2 13+125

3 13-1211

4 13-125