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A chord of parabola subtends a right angle at the vertex. Find the locus of the point of intersection of tangents at its extremities.
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Line passing through points (1/2, 1) would be: (y – 1) = m (x – 1/2) y = mx – m/2 + 1............(1) 2x + y = 3 …............(2) sove gthe eqns. (1) and (2). get x and y values in terms of m. U would ...
let the coordinates of end points for the focal chord drawn of the parabola be: P(at1^2,2at) and Q (at2^2, 2at2) respectively. SP = a+ at1^2 = > a (1+ t1^2) SQ = a(1 + t2^2) Now write SQ in terms of...
let the eqn. of the line is: x/a + y/b = 1 now pass it through (2,2) coordinates. 2/a + 2/b = 1......(1) and given a+b = 1.............(2) solve these two eqns. and get the valiues of a and b.
Solve both eqns simply and take out (x,y) coordinates. put y = mx +1 in 3x + 4y – 9 = 0 U would get, 3x + 4(mx + 1) – 9 = 0 (3 + 4m)x = 6 x = > 6/(3 + 4m) and y = 6m/(3+4m) + 1 = > (7m +3)/(3+4m) Now ...
For this proof askiitians has already provided the detailed solution. U can check the askiitian link here: http://www.askiitians.com/iit-jee-coordinate-geometry/normal-to-a-parabola.aspx
Direction cosines are actually the cosine of angles taking between the line and the axes. and the sum of square of the cosines always equals to 1.
From Famile of circles, Apply S1 + KL = 0 Where S1 = x^2 + y^2 - 2x - 2ay - 8 = 0 L = x + 2y + 5 = 0 and then put y = 0.
Simply write the eqn. of tangent for parabola y^2 = 4ax in parametric form with t1 and t2 respectively. And then solve them . that is to say that: t1y = x + at1^2 …..................(1) and t2y = x + ...
Dear Piyush, Sorry, But Your question is incomplete or Your attachement is missing. plz check again and re-upload re-post your question. Thankyou!
The question is incomplete though i am assuming that ABCD is any quadilateral. then A + B + C + D = 180 write B + C = 180 – (A + D) divide the whole eqn. by 2. (B+C)/2 = 90 – (A+D)/2.........(1) In...
Draw the suitable diagram for the given question and U would automqtically knw what do next. After drawing join the line PQ which also subtends 45 degree angle. Now try to recall pythogorus formulae...
put y = 2-x into the circle eqn. The quadratic formed would must possess the D less than zero that is to say that their roots should be imaginary for not touching the circles either way. thus U would ...
the incentre of a triangle formed by x+y=1,x=1,y=1 is the incomplete one. Plz provide the full information regarding your question and aslk again. Thankyou!
Just apply the formulae as T=S1. that is: xx1 + yy1 – a^2 = x1^2 + y1^2 – a^2 where x1 = p and y1 =q.
The inforrmation for this question is insufficient. same radii circles should align somewhere. They are intersecting non – intesrecting their position are all not given. Check the qsnt. plz again!
Assume that point in parametric form on line 2x + y = 1 as: (x , 1-2x) and apply chords of contact concept here that is: T = 0. and put the point (x,1-2x). and try to satisfy the options sequentially.
Apply Faamily of circles concept here. that is : S1 + kS2 = 0 and then for the concylic concept it must be a circle. therfore coeff. of xy should be zero and coeff. of x^2 and coeff. of y^2 should be ...
Use all the informations given and follow the step line wise. First find the S2 circle eqn. For that first find the image of center of circle S1 along y=x. U can do this by solving the eqns y=x and...
First find out the radius of the circle given: x^2 + y^2 – 2x = 0 . And assume locus of center be P(a,b) and T be touching point and And foot of a perpendicular line from P to H. then by geomtery the ...
Differntiate y^2= 4ax, 2y.dy/dx = 4a dy/dx = 2a/y dy/dx = 2a/-2a = -1 Thus for x^2 = 4ay the slope should also come to -1 as they both have a common tangent. Now doing, 2x = 4ady/dx2 dy/dx2 = x/2a =...
Equation of chord of contact from p(h,k) to x 2 + y 2 = a 2 is xh + yk = a 2 this line touches (x) 2 + (y-a) 2 = a 2 substituting x= (a 2 -yk)/h from the first equation, we get a quadratic in y.The...
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