If the normal to the ellipse `3x^(2)+4y^(2)=12` at a point P on it is parallel to the line , `2x+y=4` and the tangent to the ellipse at P passes through Q (4,4) then Pq is equal toA. `(5 sqrt(5))/(2)`B. `( sqrt(61))/(2)`C. `( sqrt(221))/(2)`D. `( sqrt(157))/(2)`

If the normal to the ellipse `3x^(2)+4y^(2)=12` at a point P on it is

1.	If the normal to the ellipse `3x^(2)+4y^(2)=12` at a point P on it is parallel to the line , `2x+y=4` and the tangent to the ellipse at P passes through Q (4,4) then Pq is equal toA. `(5 sqrt(5))/(2)`B. `( sqrt(61))/(2)`C. `( sqrt(221))/(2)`D. `( sqrt(157))/(2)`
Answer» Correct Answer - A Key idea equation of tangent and normal to the `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` at point `p(x_(V)Y_(1))` is ` T=0 =rarr (xx_(1))/(a^(2))+(y y_(1))/(b^(2))=1 and (a^(2)x)/(x_(1))-(b^(2)y)/(y_(1))=a^(2)-b^(2)` respectively . Equation of given ellipse is `3x^(2)+4y^(2)=12` `implies (x^(2))/(4)+(y^(2))/(3)=1` Now . Let point `P(2 cos theta , sqrt(3) sin theta),` so equation of tangent to ellipse (i) at point P is `( x cos theta )/(2) +( y sin theta)/(sqrt(3))=1` since, tangent (ii) passes through point Q(4,4) ` therefore 2 cos theta +(4)/(sqrt(3))=1` and equation of normal to ellipse (i) nad point P is `(4x)/(2 cos theta)-(3y)/(sqrt(3)sin theta)=4-3` `implies 2x sin theta- sqrt(3) cos theta y = sin theta cos theta` since normal (iv) is parallel to line , 2x + y =4 `therefore ` Slope normal (iv) = slope of line `, 2x+y=4` `implies (2) /(sqrt(3)) tan theta =- 2implies tan theta =-sqrt(3) implies theta = 120^(@) ` ` implies ( sin theta , cos theta)=( (sqrt(3))/(2),-(1)/(2))` hence , point `p(-1,(3)/(2))` Now `PQ=sqrt((4+1)^(2)+(4-(3)/(2))^(2))` [ given cordinates of Q`-=`(4,4)] `=sqrt(25+(25)/(4))=(5sqrt(5))/(2)`

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If the normal to the ellipse `3x^(2)+4y^(2)=12` at a point P on it is parallel to the line , `2x+y=4` and the tangent to the ellipse at P passes through Q (4,4) then Pq is equal toA. `(5 sqrt(5))/(2)`B. `( sqrt(61))/(2)`C. `( sqrt(221))/(2)`D. `( sqrt(157))/(2)`

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