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S1 :  Vertex of a parabola bisects the subtangent.

S2 :  Subnormal of a parabola is equal to its latusrectum.

S3 : Circle with focal radius of a point on parabola as diameter touches the tangent drawn at the vertex of  the parabola.

S4 : Directrix of a parabola is the tangent of a circle drawn its focal chord as diameter.

Option: 1

FTTT


Option: 2

FFTT

 


Option: 3

TTTT


Option: 4

TFTT


Answers (1)

best_answer

 

Equation of a circle in diametric form -

\left ( x-x_{1} \right )\left ( x-x_{2} \right )+\left ( y-y_{1} \right )\left ( y-y_{2} \right )= 0

 

- wherein

Where A\left ( x_{1},y_{1} \right )\, and \:B \left ( x_{2},y_{2} \right ) are the two diametric ends.

Standard equation of parabola -

y^{2}=4ax

- wherein

 

 

Equation of tangent -

yy_{1}= 2a\left ( x+x_{1} \right )

 

- wherein

Tangent at  P\left ( x,y_{1} \right )on  y^{2}=4ax

S1 : Tangent at A(at^{2},2at) is y=\frac{x}{t}+at

This intersect the x-axis at\:\:(-at^{2},0) and foot of \perp from A on the x-axis is (at^{2},0) clearly origin is the mid point.

\therefore S1 is true.

S2 : Equation of normal at \:\:(at^{2},2at) is y=-t\: x+2\: at+at^{3}

it intersect x-axis at (2a+at^{2},0)

\therefore   subnormal =2a

\therefore S2 is False.

S3 : Let A (at^{2}, 2at) be a point on the parabola y^{2} = 4ax. S (a, 0) be the focus equation of circle having AS as diameter is (x - at^{2}) (x - a) +y (y - 2at) = 0 and tangent at the vertex to the parabola is x = 0. It can be easily checked that x = 0 touches this circle.

\therefore S3 is true by D=0

S4 : equation of such circle is

(x-at^{2})\left ( x-\frac{a}{t^{2}} \right )+(y-2at)\left ( y+\frac{2a}{t} \right )=0

Directrix x=-a which is tangent.

\therefore S4 is true.

 

Posted by

Anam Khan

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