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1. Find the point P on the line 2x+3y+1=0 such that |PA-PB| is maximum, where

A(2,0) and B(0,2).

Ans:- (7,-5)

In the PAB, AB > |PA-PB| (Triangle Inequality). And AB will be equal

to |PA-PB| if P lies on the straight line AB. Hence |PA-PB| is maximum whenP lies on the straight line AB.

Equation of AB: x+y-2=0.

Hence by solving equation of AB and the given straight line 2x+3y+1=0, we

get P as (7,-5).

2. The orthocentre of the triangle formed by the lines xy=0 and x+y=1 is _____ ..

Ans:- (0,0)

xy=0, gives rise to x=0 and y=0. And the third side is x+y=1. Hence the

triangle formed by the sides is right angled triangle, with the right angle at

(0,0). Orthocentre of any right angle lies on the vertex, where right angle is

formed. Hence answer is origin (0,0).

3. The co-ordinates of the point from which the length of the tangents to the

three circles x2 + y2 = 1, x2+y2+8x+15=0, x2+y2+10y+24=0 are equal, is _____.

Ans:- (-2,-2.5)

From radical centre of 3 circles, the lengths of the tangents are all

equal. Radical centre can be found out by finding out the intersection point of

two radical-axes of two circles taken in pairs.

Radical Axis of C1 & C2: (x2 + y2 1) (x2+y2+8x+15) =0 or 8x+16=0 or x=-2

Radical Axis of C1 & C3: (x2 + y2 1) (x2+y2+10y+24) =0 or 10y+25=0 or y=-

2.5Hence the required point is (-2,-2.5)

4. A point moves so that square of its distance from the point (3,-2) is

numerically equal to its distance from the line 5x-12y=13. Find the equation to

its locus.

Ans:- 13(x2+y2) -83x + 64y +182=0 if 5x-12y-130 &

13(x2+y2) -73x + 40y +156=0 if 5x-12y-13

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5. A line APB of constant length meets the x-axis at A and y-axis at B. If AP=a,

PB=b and the line slides with the extremities on the co-ordinate axes, Show

that the equation of the locus of the point P is

Ans:- Let the co-ordinates of any point on the locus be (h,k).In the figure PBB ~ APA BP/AA=BP/PA

Or h/( )=b/a By simplifying we get

Hence the equation of the locus of the point P is

6. If the line 2x+3y=1 touches the parabola y2=4ax at the point P, then the focal

distance of the point P is ____ .

Ans:- 13/18

Equation of the tangent to the parabola y2=4ax at (at2,2at)will be of the

form ty=x+at2.

Equation of the given line can be re-arranged as 3y=-2x + 1

By comparing both we get,

t = -3/2 & a =-2/9Focal distance of the point = x-co-ordinate + a = at2 + a = 13/18

7. Find the Centres & Radii of the circles which touch the straight lines x+y=2, x-

y=2 and also touches the circle x2+y2= 1

Ans:- Centre (2,0) Radius 2-1 & Centre (-4-32,0) Radius 3+32As the circles touch x+y =2 & x-y=2, the centre of the circles lie on the

angle bisectors of these straight lines, which are x-axis(y=0) & x=2. But x=2 is

not possible as the circle also touches the circle x2+y2= 1.

Let the co-ordinates of circle having centre on y=0 is (a,0)

Radius of this circle will be |a-2|/2This touches the circle x2+y2=1 the distance from origin is equal to sum of itsradius and the radius of the given circle(1).

So we can write |a| = 1 + |a-2|/2

There are 2 conditions, if a0 or a

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Ans:- The common chord of the two circles 2x(g-g) + 2y(f-f) + c-c =0.

For the first circle to bisect the circumference of the second circle, the

common chord has to be the diameter, hence passing through the centre of

the second circle (-g,-f).

So the required condition is -2g(g-g) - 2f(f-f) + c-c =0Or 2g(g-g) + 2f(f-f)=c-c

10. Find the area enclosed by the four lines ax by c = 0.

Ans:- 2c2/ab

The four sides are ax+by+c=0, ax+by-c=0, ax-by+c=0 & ax-by-c=0.

Clearly the straight lines are forming a parallelogram (1st two straight lines

being parallel to each other & 2nd two being parallel to each other.)

Distance between First two sets of straight lines & 2nd two sets of straight lines

= d=2|c|/(a2+b2)0.5

Hence we know that, these 4 straight lines form a rhombus.

Slope of those straight lines are a/b & -a/b

Angle between those two straight lines = = tan-1

= tan-1

So sin=2ab/(a2+b2)So the area of the rhombus = (d/sin).d = d2 /sin= 2c2/ab