Given parabola is y2=32x, So, 4a=32 ⇒‌‌a=8 The point P is (8,16). This in parametric form as (at2,2at) So, 8t2=8 ⇒‌‌t2=1⇒t=±1 And, 2at=2(8)t=16t=16 ⇒‌‌t=1 So, the perimeter for P is t1=1. The equation of the normal to the parabola at a point P(at12,2at1) is ‌y+t1x=2at1+at13 ‌⇒‌‌y+(1)x=2(8)(1)+8(1)3 ‌⇒‌‌y+x=16+8=24 ‌⇒‌‌y=24−x Substitute this into parabola equation ‌y2=32x ‌⇒‌‌(24−x)2=32x ‌⇒‌‌576−48x+x2=32x ‌⇒‌‌x2−80x+576=0 Since, x=8 is one root corresponding to point P. Let the other root be xQ. Using the sum of roots, 8+xQ=80 ⇒xQ=72 So, y=24−xQ =24−72=−48 So, the coordinates of Q are (72,−48) Now, the equation of the tangent to parabola y2=4ax at a point (x1,y1) is ‌yy1=2a(x+x1) ‌⇒y(−48)=2(8)(x+72)=16(x+72 ‌⇒−3y=x+72⇒x+3y+72=0 So, the equation of the tangent drawn at Q to the parabola is x+3y+72=0