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1.

A point moves so that the sum of its distances from `(a e ,0)a n d(-a e ,0)`is `2a ,`prove that the equation to its locus is `(x^2)/(a^2)+(y^2)/(b^2)=1`, where `b^2=a^2(1-e^2)dot`

Answer» AP+BP=2h
`AP=sqrt((h-he)^2+k^2)`
`BP=sqrt((h+he)^2+k^2`
`AP+BP=2h`
`(AP+BP)^2=4a^2`
`AP^2+BP^2+2(AP)(BP)=4a^2`
`(h-ae)^2+k^2+(h+ae)^2+k^2+2sqrt(((h-ae)^2+k^2)(h+ae)^2+k^2`
`h^2/a^2+k^2/b^2=1`
Putting h=x,k=y
`x^2/a^2+y^2/b^2=1`.
Discussion
2.

Find the point to which the origin should be shifted so that theequation `y^2-6y-4x+13=0`is transformed to the form `y^2+A x=0.`

Answer» `y^2+Ax=0`
`y^2-6y+(3^2-3^2)-4x+13=0`
`(y-3)^2+4-4x=0`
`(y-3)^2+4(1-x)=0`
`(y-3)^2-4(x-1)=0`
`y^2+Ax=0`
`A=-4`.`y-3=0`
`y=3`
`x-1=0`
`x=1`
Point(1,3).
Discussion
3.

If the axes are shifted to the point `(1,-2)`without rotation, what do the following equations become?`2x^2+y^2-4x+4y=0``y^2-4x+4y+8=0`

Answer» First equation
`2x^2+y^2-4x+4y=0`
`2(x-1)^2+(y+2)^2-4(x-1)+4(y+2)=0`
`2(x^2+1-2x)+y^2+4+4y-4x+4+4y+8=0`
`2x^2+2-2x+y^2+4-4y-4x+4+4y+8=0`
`2x^2+y^2-6x+8y+18=0`
`y^2-4x+4y+8=0`
`y^2+8y-4x+24=0`.
Discussion
4.

Find the locus of a point, such that the join of `(-5,1)a n d(3,2)`subtends a right angle at the moving point.

Answer» `y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)`
Equation of AB
`y-1=(2-1)/(3+5)(x+5)`
`y-1=1/8(x+5)`
`m=1/8`
Slope of PD=-8
at point(h,k)
`y-k=-8(x-h)`.
Discussion
5.

Find the equation of the locus of a point which moves such that theratio of its distances from `(2,0)a n d(1,3)`is `5: 4.`

Answer» `(AP)/(BP)=5/4`
`(AP)^2/(BP)^2=(25/16)``d=sqrt((x_1-x_2)^2+(y_1-y_2)^2`
`d^2=(x_1-x_2)^2+(y_1-y_2)^2`
`(AP)^2=(h-2)^2+(k-0)`
`(BP)^2=(h-1)^2+(k-3)^2`
`(AP)^2/(BP)^2=25/16`
`((h-2)^2+k^2)/((h-1)^2+(k-3)^2)=25/16`
`16h^2+64-64h+16k^2=25h^2+25-50h+25k^2+225-150h`
`9h^2+9k^2+14h-150k+186=0`
Putting h=x,k=y
`9x^2+9y^2+14x-150y+186=0`.
Discussion
6.

`A(5,3), B(3,-2)`are two fixed points; find the equation to the locus of a point `P`which moves so that the area of the triangle `P A B`is 9 units.

Answer» area of triangle`=1/2|[x_!,y_1,1],[x_2,y_2,1],[x_3,y_3,1]|`
`/_=1/2|[h,k,1],[5,3,1],[3,-2,1]|=9`
`/_=h(3+2)-k(5-3)+1(-10-9)=18`
`5h-2k-19=18`
`5h-2k=37`
putting h=x,k=y
`5x-2y=37`.
Discussion