Question

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Question : F1 and F2 are two points situated at (3 , 0) and (–3 , 0) respectively. If P (x , y) is a point on the curve 16x² + 25y² = 400, then PF1 + PF2 is equal to.....? (JEE Mains)​

Answer 1

Solution

$\implies \: \sf {x}^2/25 \: {y}^2 \: 16 = 1$

Here,

$\implies \: \sf {a}^2 = 25, {b}^2 = 16$

but,

$\implies \: \sf {b}^2 = {a}^2 ( 1 - {e}^2)$

$\sf \implies 16 = 25 ( 1 - {e}^2 )$

$\sf \implies 16/25 = 1 - {e}^2$

$\sf \implies{e}^2 = 1 - 16/25 = 9 / 25 \\ \sf \implies e = 3/5$

We have,

$\sf 3 = a 3/5 // a = 5$

Now, PF 1 + PF 2 = major axis = 2a

= 2 × 5 = 10

Answer 2

EXPLANATION.$\sf : \implies \: f_{1} \: and \: f_{2}\: are \: two \: points \: situated \: at \: (3 ,0) \: \: and \: ( - 3 ,0) \\ \\ \sf : \implies \: p \: (x,y) \: is \: a \: point \: on \: curve \: 16 {x}^{2} + 25 {y}^{2} = 400$

$\sf : \implies \: \:{ \underline {\underline{to \: find \: pf_{1} \: + p f_{2} }}}$$\sf : \implies \: general \: equation \: of \: an \: ellipse \\ \\ \sf : \implies \: \frac{ {x}^{2} }{ {a}^{2} } + \: \frac{ {y}^{2} }{ {b}^{2} } = 1 \\ \\ \sf : \implies \: 16 {x}^{2} + 25{y}^{2} = 400 \\ \\ \sf : \implies \: \frac{16 {x}^{2} }{400} + \frac{25 {y}^{2} }{400} = 1 \\ \\ \sf : \implies \: \frac{ {x}^{2} }{25} + \frac{ {y}^{2} }{16} = 1$

$\sf : \implies \: {a}^{2} = 25 \: \: and \: \: \: {b}^{2} = 16 \\ \\ \sf : \implies \: eccentricity \: of \: an \: ellipse \\ \\ \sf : \implies \: {b}^{2} = {a}^{2} (1 - {e}^{2} ) \\ \\ \sf : \implies \: 16 = 25(1 - {e}^{2} ) \\ \\ \sf : \implies \: 16 = 25 - 25 {e}^{2} \\ \\ \sf : \implies \: 9 = 25 {e}^{2}$

$\sf : \implies \: e \: = \sqrt{ \dfrac{9}{25} } = \dfrac{3}{5} \\ \\ \sf : \implies \: eccentricity \: = \frac{3}{5} \\ \\ \sf : \implies \: foci \: of \: an \: ellipse \: = (ae,0) \: \: and \: \: ( - ae,0) \\ \\ \sf : \implies \: (5 \times \frac{3}{5} ,0) \: \: and \: \: ( - 5 \times \frac{3}{5} ,0) \\ \\ \sf : \implies \: (3,0) \: \: and \: ( - 3,0)$

$\sf : \implies \: let \: p \: is \: a \: point \: on \: ellipse \: \\ \\\sf : \implies \: distance \: of \: foci \: from \:\: point \: p \: is \\ \\ \sf : \implies \: pf_{1} \: + \: p f_{2} \: =( a \: + \: ae) + (a - ae) \\ \\ \sf : \implies \: pf_{1} \: + \: p f_{2} \: \: = 2a = 2 \times 5 = 10$

$\sf : \implies \: \green{\underline{ \underline{ pf_{1} \: + \: p f_{2} \: = 2a \: = 2 \times 5 = 10}}}$