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51.

Let P be any point on a directrix of an ellipse of eccentricity e, S be the corresponding focus and C be the centre of the ellipse. The line PC meets the ellipse at A. The angle between PS and tangent at A is `alpha`. Then `alpha` is equal to

Answer» `m_(PS)=(K-0)/((a/e)-ae)`
`=K/(a((1-e^2)/e)`
`=(Ke)/(a(1-e^2))`
`(x x_1)/a^2+(yy_1)/b^2=1`
`m_T=-x_1/a^2*b^2/y_1`
`=x_1/y_1*(a^2(1-e^2))/a^2`
`Ax+By+C=0`
`m=-(A/B)`
`=-x_1/y_1(1-e^2)`
`m_T=-a/(Ke)(1-e^2)`
`m_(PS)=(Ke)/(a(1-e^2))`
`m_T=(-a(1-e^2))/(Ke)`
`m_(PS)*m_(T)=-1`
PS is perpendicular to tangent.
Option B is correct.
Discussion
52.

Find an equation of the circle with centre at (0,0) and radius r.

Answer» the equation of circle is
`(x-a)^2 + (y-b)^2 = r^2`
`(a,b) `is centre given as `(0,0)`
putting in the equation
`x^2 + y^2 = r^2`
answer
Discussion
53.

Find the centre and the radius of the circle `x^2+y^2+8x+10 y-8=0`.

Answer» `x^2 + y^2 + 8x + 10y - 8 =0`
`x^2 + 8x + 16 -16 + y^2 + 10y +25- 25 - 8=0`
`(x+4)^2 - 16 + (y+ 5)^2 - 25 -8=0`
`(x+4)^2 + (y+5)^2 = 33+ 16`
`(x+4)^2 + (y+5)^2 = (7)^2`
so centre is `(h,k) = (-4,-5)`
radius `= 7`units
answer
Discussion
54.

Find the equation of tangent to the conic `x^2-y^2-8x+2y+11=0` at `(2,1)`

Answer» Equation of the conic is ,
`x^2-y^2-8x+2y+11=0`
Differentiating it with respect to `x`,
`2x-2ydy/dx-8+2dy/dx = 0`
`=>dy/dx(2-2y) = 8-2x`
`=>dy/dx = (4-x)/(1-y)`
At point `(2,1)`,
`dy/dx = oo`
`:.` slope is `oo`.Now, equation of line,`y = mx+c`
At point `(2,1)`,
`1 = 2m+c => c = 1-2m`
`:.` Equation of tangent,
`y = m(x)+(1-2m)`
`=>y = m(x-2)+1`
`=>(y-1)/m = (x-2)`
As `m = oo`,
`:. x-2 = 0`
`=>x=2`, which is the required equation.
Discussion
55.

The locus of the point of intersection of lines `sqrt3x-y-4sqrt(3k)`=0 and `sqrt3kx+ky-4sqrt3=0` for different value of k is a hyperbola whose eccentricity is 2.

Answer» True
Given equation of line are
`sqrt3x-y-4sqrt3k`=0 …(i)
and `sqrt3kx+ky-4sqrt3=0`
From Eq. (i) `4sqrt3k=sqrt3x-y`
`rArr k=(sqrt3x-y)/(4sqrt3)`put in Eq. (ii), we get
`sqrt3x((sqrt3x-y)/(4sqrt3))+((sqrt3x-y)/(4sqrt3))y-4sqrt3=0`
`rArr1/4(sqrt3x^(2)-xy)+1/4(xy-(y^(2))/(sqrt3))-4sqrt3=0`
`rArr (sqrt3)/4x^(2)-(y^(2))/(4sqrt3)-4sqrt3=0`
`rArr 3x^(2)-y^(2)-48=0`
`rArr 3x^(2)-y^(2)=48`,which is hyperbola.
Discussion
56.

The locus of the point of intersection of tangents drawn at the extremities of normal chords to hyperbola `xy=c^2` is (A) `(x^2-y^2 )^2 + 4c^2xy = 0` (B)` (x^2+y^2)^2+ 4c2^xy=0` (C)` x^2-y^2 )^2 + 4cxy = 0` (D) `(x^2 +y^2)^2+4cxy =0`

Answer» Equation of tangent to hyperbola `xy=c^2` is:
`x(x1)+y(y1)=2c^2 => (1)`Equation of normal chord at point(h,K) will be:
`hx-ky=h^2-k^2 => (2)`On equating (1) and (2), we get
`h/(x1)=-k/(y1)=(h^2-k^2)/(2c^2)`
On solving,The locus is:
`(x^2-y^2)^2+4c^2xy=0`
Discussion
57.

The eccentricity of the hyperbola whose length of the latus rectum isequal to 8 and the length of its conjugate axis is equal to half of thedistance between its foci, is :A. `4/3`B. `4/sqrt3`C. `2/sqrt3`D. none of these

Answer» Correct Answer - C
Given that, `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 altb`
We know that, `e=sqrt(1-(a^(2))/(b^(2)))rArre^(2)=((b^(2)-a^(2)))/(b^(2))`
`rArr b^(2)e^(2)=b^(2)=a^(2)`
`rArra^(2)=b^(2)(1-e^(2))`
Discussion
58.

Find the eccentricity of the hyperbola `9y^(2)-4x^(2)=36`

Answer» Given equation of the hyperbola is
`9y^(2)-4x^(2)=36`
`rArr (9y^(2))/36-(4x^(2))/36=36/36`
`rArr (y^(2))/4-(x^(2))/9=1`
`rArr -(x^(2))/9+(y^(2))/4=1`
Since, this equation in form of `-(x^(2))/a+(y^(2))/(b^(2))=1`.wherea=3 and b=2.
`e=sqrt(1+(a/b)^(2))`
`=sqrt(1+9/4)=(sqrt13)/2`
Discussion