Parabolas from Focus and Directrix, and Tangent Lines to a Circle: Geometry Math Worksheet Answer Key – PDF Download

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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