# Doubt from straight lines

Plz solve question 2.

Take a parametric coordinate in curve reflection of that about line will give other point find locus of point..

\textbf {jee advanced 2015 type problem}

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How to take the parametric coordinate??

See easy way without taking parametric also
x=0,y=0 , x=a,y=a satisfying reflect them along line u will get point nd satisfy in option u will get ans d option

Or

x=at^2, y=at^3
U may gooped up in calculation in parametric form i think so gave other approch to

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Yes I solved it the first way, but thought that it would be easier if parametric coordinates are known. But as you said it is lengthy.

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Family of lines try partial differentiation mthd

Can you pls elaborate, I'm kinda weak here.

I m sorry that may be a mthd but i hvnt tried the question is going for homogenization

y=mx+c
\dfrac{y-mx}{c }=1
Then 1 degree term in given question write this in place of 1
Now coefficient of x^2+y^2=0

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First reflect across x-axis, so the y- coordinate changes sign and we are dealing with the curves

ay^2=x^3 and x-y=a

Now let X=x-a, Y=y i.e. translate origin of axes to (a,0). Then we have to reflect aY^2 = (X+a)^3 across X-Y=0

Now, the reflected curve is aX^2 = (Y+a)^3 i.e. a(x-a)^2 = (y-a)^3

Now reflect this curve across x- axis and we get a(x-a)^2 = (a-y)^3

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#18: As suggested by @Sneha_2021 use homogenization. Take the chord as y=mx+c so that \dfrac{y-mx}{c}=1

Hence 2x^2+3y^2 - 5x \left( \dfrac{y-mx}{c} \right)=0 or (5m+2c)x^2-5xy+3cy^2=0 represents the pair of lines joining origin to the point of intersection of the chord with the ellipse.

and since the pair of lines are orthogonal we have coeff of x^2+coeff of y^2=0 i.e. 5m+5c =0 or m+c=0

Hence the equation of the chord is of the form y=m(x-1) which always passes through (1,0)

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Thank you @Hari_Shankar sir

Can anyone solve question 46

2 may be
#hint
x^2+y^2<16
Where x=\alpha ,y=\alpha+1

Now put value of \alpha hit nd trail
2 ans u will get