This geometry parabola tangents worksheet answer key solves all 24 problems of the parabolas from focus and directrix, and tangent lines to a circle worksheet step by step.
This answer key shows a full worked solution for each problem, including the distance comparisons, the algebra for every equation and the right triangles for every tangent. Check each answer against the stated check value.
Geometry Parabola Tangents Worksheet Answer Key: problems 1–5
Problem 1: Checking a point against the definition
(a) Distance to the focus: \(\sqrt{4^{2}+(4-1)^{2}}=\sqrt{25}=5\). Distance to the directrix: \(4-(-1)=5\). The distances are equal, so A is on the parabola.
(b) Distance to the focus: \(\sqrt{6^{2}+7^{2}}=\sqrt{85}\approx 9.22\). Distance to the directrix: \(8+1=9\). They differ, so B is not on it. Answer: A is on the parabola; B is not
Problem 2: Solving for a missing ordinate
The distance to the directrix is \(8+3=11\). The distance to the focus is \(\sqrt{(8-3)^{2}+y^{2}}=\sqrt{25+y^{2}}\).
Set \(\sqrt{25+y^{2}}=11\): \(25+y^{2}=121\), so \(y^{2}=96\) and \(y=\pm 4\sqrt{6}\approx\pm 9.80\). Check with \(y^{2}=12x=96\). Answer: \(y\approx 9.80\) or \(y\approx -9.80\)
Problem 3: Deriving an equation from the definition
For a point \((x,y)\): \(\sqrt{x^{2}+(y-4)^{2}}=|y+4|\).
Square both sides: \(x^{2}+y^{2}-8y+16=y^{2}+8y+16\). Subtract \(y^{2}+16\) from both sides and add \(8y\): \(x^{2}=16y\). Answer: \(x^{2}=16y\)
Problem 4: A parabola that opens downward
The vertex is the midpoint \((0,0)\), and the focus is below it, so \(p=-2\). Then \(x^{2}=4py=4(-2)y\).
Check with the point \((4,-2)\): distance to the focus \(4\), distance to the directrix \(|-2-2|=4\), and \(4^{2}=16=-8(-2)\). Answer: \(x^{2}=-8y\)
Problem 5: A parabola that opens to the right
Set \(\sqrt{(x-3)^{2}+y^{2}}=|x+3|\) and square: \(x^{2}-6x+9+y^{2}=x^{2}+6x+9\). Then \(y^{2}=12x\), which matches \(y^{2}=4px\) with \(p=3\). Answer: \(y^{2}=12x\)
Geometry Parabola Tangents Worksheet Answer Key: problems 6–10
Problem 6: Focus and directrix from an equation
Compare with \(x^{2}=4py\): \(4p=20\), so \(p=5\). The focus is \((0,p)=(0,5)\) and the directrix is \(y=-p=-5\). Answer: focus \((0,5)\); directrix \(y=-5\)
Problem 7: Rewriting a quadratic function
Multiply by 8: \(x^{2}=8y\). Then \(4p=8\) and \(p=2\). The focus is \((0,2)\) and the directrix is \(y=-2\). Answer: focus \((0,2)\); directrix \(y=-2\)
Problem 8: A sideways parabola
Compare with \(y^{2}=4px\): \(4p=-16\), so \(p=-4\). The focus is \((p,0)=(-4,0)\) and the directrix is \(x=-p=4\). Since \(p < 0\), it opens to the left. Answer: focus \((-4,0)\); directrix \(x=4\); opens left
Problem 9: Vertex away from the origin
The vertex is the midpoint of \((3,6)\) and the point \((3,2)\) on the directrix: \((3,4)\). The focus is 2 above it, so \(p=2\).
Equation: \((x-3)^{2}=4(2)(y-4)=8(y-4)\).
Check \((7,6)\): \((7-3)^{2}=16\) and \(8(6-4)=16\). Distances: to the focus \(4\), to the directrix \(6-2=4\). Answer: vertex \((3,4)\), \(p=2\), \((x-3)^{2}=8(y-4)\)
Problem 10: Reading a shifted equation
Compare with \((x-h)^{2}=4p(y-k)\): \(h=-2\), \(k=1\), \(4p=12\), \(p=3\).
Vertex \((-2,1)\); focus \((h,k+p)=(-2,4)\); directrix \(y=k-p=-2\). Answer: vertex \((-2,1)\); focus \((-2,4)\); directrix \(y=-2\)
Geometry Parabola Tangents Worksheet Answer Key: problems 11–15
Problem 11: Focus and a vertical directrix
The vertex is halfway between \(x=-1\) and \(x=5\) at \(y=3\): \((2,3)\). The focus is 3 units to the left of the vertex, so \(p=-3\).
\((y-3)^{2}=4(-3)(x-2)=-12(x-2)\). Check the point \((-1,9)\): \((9-3)^{2}=36=-12(-3)\), and the distances to the focus and to \(x=5\) are both 6. Answer: \((y-3)^{2}=-12(x-2)\)
Problem 12: Completing the square
Isolate the x-terms: \(x^{2}-6x=8y-1\). Add 9 to both sides: \((x-3)^{2}=8y+8=8(y+1)\).
So \(h=3\), \(k=-1\), \(p=2\). Vertex \((3,-1)\); focus \((3,1)\); directrix \(y=-3\). Answer: \((x-3)^{2}=8(y+1)\); vertex \((3,-1)\); focus \((3,1)\); directrix \(y=-3\)
Problem 13: Tangent perpendicular to a radius
The tangent is perpendicular to the radius, so triangle OTP has a right angle at T: \(OP^{2}=5^{2}+12^{2}=169\), so \(OP=13\). Answer: \(OP=13\)
Problem 14: Length of a tangent segment
The tangent segment is a leg of a right triangle with hypotenuse \(OP=17\) and other leg 8: \(PT=\sqrt{17^{2}-8^{2}}=\sqrt{225}=15\). Answer: 15 units
Problem 15: Finding the radius
\(r^{2}+24^{2}=25^{2}\), so \(r^{2}=625-576=49\) and \(r=7\). Answer: 7
Geometry Parabola Tangents Worksheet Answer Key: problems 16–20
Problem 16: Equal tangent segments
Tangent segments from one point are congruent: \(3x+4=5x-6\). Then \(10=2x\), so \(x=5\).
\(PA=3(5)+4=19\) and \(PB=5(5)-6=19\). Answer: \(x=5\); \(PA=19\)
Problem 17: Angles around two tangents
The radii \(OA\) and \(OB\) are perpendicular to the tangents, so \(\angle OAP=\angle OBP=90^{\circ}\). The angles of quadrilateral OAPB add to \(360^{\circ}\):
\(m\angle APB=360^{\circ}-90^{\circ}-90^{\circ}-110^{\circ}=70^{\circ}\). Answer: \(70^{\circ}\)
Problem 18: Is the line tangent?
A line is tangent at T exactly when \(OT\perp PT\), so test the Pythagorean relationship.
(a) \(9^{2}+12^{2}=81+144=225=15^{2}\): right angle at T, so tangent.
(b) \(7^{2}+24^{2}=49+576=625\), but \(26^{2}=676\): no right angle, so not tangent. Answer: (a) tangent; (b) not tangent
Problem 19: Tangent line in the coordinate plane
The radius from \(O(0,0)\) to \(T\) has slope \(\frac{4}{3}\), so the tangent has slope \(-\frac{3}{4}\).
\(y-4=-\frac{3}{4}(x-3)\), so \(4y-16=-3x+9\) and \(3x+4y=25\).
At \(y=0\): \(x=\frac{25}{3}\). Answer: \(3x+4y=25\); crosses the x-axis at \((\frac{25}{3},0)\)
Problem 20: Why the midpoint circle works
(a) \(MO=\frac{1}{2}\cdot 10=5\).
(b) \(OP\) is a diameter of the dashed circle and A is on that circle, so \(\angle OAP\) is inscribed in a semicircle and equals \(90^{\circ}\). Then \(PA\perp OA\), so PA is tangent.
(c) \(PA=\sqrt{10^{2}-6^{2}}=\sqrt{64}=8\). Answer: \(MO=5\); right angle by the semicircle; \(PA=8\)
Geometry Parabola Tangents Worksheet Answer Key: problems 21–24
Problem 21: Cable to a water tank
The cable is a tangent segment, perpendicular to the radius at the touching point: \(\sqrt{20^{2}-12^{2}}=\sqrt{256}=16\). Answer: 16 ft
Problem 22: Satellite dish
(a) \(p=1.5\), so \(x^{2}=4(1.5)y=6y\).
(b) At the rim \(x=4\): \(y=\frac{16}{6}=\frac{8}{3}\approx 2.67\) ft.
(c) The rim is \(\frac{8}{3}+\frac{3}{2}=\frac{25}{6}\) ft from the directrix, hence from the focus: \(\sqrt{16+(\frac{8}{3}-\frac{3}{2})^{2}}=\sqrt{\frac{625}{36}}=\frac{25}{6}\approx 4.17\) ft. Answer: \(x^{2}=6y\); about 2.67 ft; about 4.17 ft
Problem 23: Headlight reflector
(a) \(p=2.5\), so \(x^{2}=10y\).
(b) At \(x=5\): \(y=\frac{25}{10}=2.5\) in.
(c) The rim is at \((5,2.5)\) and the bulb at \((0,2.5)\), 5 in. apart. Check: the rim is \(2.5+2.5=5\) in. from the directrix. Answer: \(x^{2}=10y\); 2.5 in.; 5 in.
Problem 24: Tangents from a point on the axis
(a) \(OP=5\), \(r=4\): \(PT=\sqrt{25-16}=3\).
(b) For a point \(T(x,y)\) of tangency, \(OT\perp PT\): \(x(x-5)+y^{2}=0\). With \(x^{2}+y^{2}=16\) this gives \(16-5x=0\), so \(x=3.2\) and \(y^{2}=16-10.24=5.76\), \(y=\pm 2.4\). Check: \(\sqrt{(5-3.2)^{2}+2.4^{2}}=3\).
(c) Slope of \(PT\): \(\frac{2.4-0}{3.2-5}=-\frac{4}{3}\). Then \(y=-\frac{4}{3}(x-5)\), or \(4x+3y=20\). Check \(T\): \(12.8+7.2=20\). Answer: 3; \((3.2,\ 2.4)\) and \((3.2,\ -2.4)\); \(4x+3y=20\)
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